Logarithmic Differentiation 5.5 of NCERT Class 12 Maths Chapter 5 introduces a powerful technique of differentiation using logarithm. This method is particularly useful for differentiating functions in which the variable appears both in the base and the exponent, such as y=u(x)^{v(x)}.
By using Logarithmic Differentiation, we will simplify the differentiation of complicated expressions involving several factors, powers and roots. Instead of repeatedly applying the Product Rule and Quotient Rule, we first take the logarithm of both sides and then differentiate implicitly. This often leads to shorter and more elegant solutions.
Key Concepts
1. Logarithmic Differentiation
Logarithmic Differentiation is a method of differentiation in which we first take the logarithm of both sides of an equation and then differentiate implicitly. For example,
If y=f(x)
then we take logarithm on both sides to obtain
\log y=\log(f(x))
This technique is particularly useful when the function contains powers, products, quotients or roots that are difficult to differentiate directly.
2. When to Use Logarithmic Differentiation
Logarithmic differentiation is especially useful in the following situations:
- Functions of the form y=u(x)^{v(x)}, where both the base and exponent contain variables.
- Functions involving many factors multiplied together.
- Functions involving quotients of complicated expressions.
- Functions containing several roots and fractional powers.
3. Important Logarithmic Properties
| S.No. | Property |
|---|---|
| 1. | \log_a(MN)=\log_aM+\log_aN |
| 2. | \log_a\left(\frac{M}{N}\right)=\log_aM-\log_aN |
| 3. | \log_a(M^n)=n\log_aM |
| 4. | \log_a1=0 |
| 5. | \log_aa=1 |
| 6. | \log_ab=\frac{\log_cb}{\log_ca} (Base Change Formula) |
4. Steps in Logarithmic Differentiation
- Take logarithm on both sides of the equation.
- Use logarithmic properties to simplify the expression.
- Differentiate both sides implicitly with respect to x.
- Substitute the value of y from the original equation.
- Simplify the final answer.
5. Functions of the Form y=u(x)^{v(x)}
Logarithmic differentiation is particularly useful for functions in which both the base and the exponent contain variables.
A general form of such a function is
y=u(x)^{v(x)}
Direct differentiation of such functions is not possible using the standard rules of differentiation.
Taking logarithm on both sides, we get
\log y=v(x)\log u(x)
We can then differentiate implicitly and finally substitute the value of y from the original equation.
Example: x^x,\;(x^2+1)^{\sin x},\;(\sin x)^x
6. Why Logarithmic Differentiation is Useful
Logarithmic differentiation often converts complicated expressions into simpler forms by using the properties of logarithms.
By converting products into sums, quotients into differences and powers into coefficients, logarithmic differentiation often makes calculations shorter and easier.
Important: Logarithmic Differentiation is a powerful technique for differentiating functions involving variable bases and variable exponents. It simplifies complicated products, quotients and powers by converting them into sums and differences using logarithmic properties, making differentiation much easier and more systematic.
Let us now solve all the NCERT questions step by step in this exercise 5.5 of Logarithmic Differentiation.
Question 1: Logarithmic Differentiation 5.5
1. Differentiate \cos x\cdot\cos 2x\cdot\cos 3x with respect to x.
Solution
Let
y=\cos x\cdot\cos 2x\cdot\cos 3x\qquad ...(1)
Taking logarithm on both sides, we get
\log y=\log(\cos x\cdot\cos 2x\cdot\cos 3x)
=\log(\cos x)+\log(\cos 2x)+\log(\cos 3x)\qquad ...(2)
(Using \log(MNP)=\log M+\log N+\log P)
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\big(\log(\cos x)\big)+\frac{d}{dx}\big(\log(\cos 2x)\big)+\frac{d}{dx}\big(\log(\cos 3x)\big)
or \displaystyle \frac{1}{y}\frac{dy}{dx}=-\tan x-2\tan 2x-3\tan 3x\qquad ...(3)
Multiplying both sides of (3) by y, we get
\displaystyle \frac{dy}{dx}=-y\big(\tan x+2\tan 2x+3\tan 3x\big)\qquad ...(4)
Using equation (1) in (4), we get
\displaystyle \frac{dy}{dx}=-\cos x\cos 2x\cos 3x\big(\tan x+2\tan 2x+3\tan 3x\big)
Hence,
\displaystyle \frac{d}{dx}\big(\cos x\cdot\cos 2x\cdot\cos 3x\big)=-\cos x\cos 2x\cos 3x\big(\tan x+2\tan 2x+3\tan 3x\big)
Question 2: Continuity and Differentiability 5.5
2. Differentiate \displaystyle \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} with respect to x.
Solution
Let
\displaystyle y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\qquad ...(1)
Taking logarithm on both sides, we get
\displaystyle \log y=\log\left[\left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{1/2}\right]
\displaystyle =\frac{1}{2}\log\left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)\qquad ...(2)
(Using \log(M^n)=n\log M)
Also,
\displaystyle \log y=\frac{1}{2}\Big[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)\Big]\qquad ...(3)
(Using \log\left(\frac{MN}{PQR}\right)=\log M+\log N-\log P-\log Q-\log R)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\qquad ...(4)
Multiplying both sides of (4) by y, we get
\displaystyle \frac{dy}{dx}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\qquad ...(5)
Hence, using equation (1) in (5), we get
\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]
You may already be following Maths Better for NCERT Solutions for the topics like
Likewise this exercise of Continuity and Differentiability 5.5 for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s move on to the next question of Exercise 5.5.
Question 3: Logarithmic Differentiation 5.5
3. Differentiate (\log x)^{\cos x} with respect to x.
Solution
Let
y=(\log x)^{\cos x}\qquad ...(1)
Taking logarithm on both sides, we get
\log y=\log\Big((\log x)^{\cos x}\Big)
=\cos x\cdot\log(\log x)\qquad ...(2)
(Using \log(M^n)=n\log M)
Using the Product Rule,
\displaystyle \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=\cos x\cdot\frac{d}{dx}\big(\log(\log x)\big)+\log(\log x)\cdot\frac{d}{dx}(\cos x)
Therefore,
\displaystyle \frac{1}{y}\frac{dy}{dx}=\cos x\left(\frac{1}{x\log x}\right)-\sin x\,\log(\log x)\qquad ...(3)
Multiplying both sides of (3) by y, we get
\displaystyle \frac{dy}{dx}=y\left[\frac{\cos x}{x\log x}-\sin x\,\log(\log x)\right]\qquad ...(4)
Using equation (1) in (4), we get
\displaystyle \frac{dy}{dx}=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\,\log(\log x)\right]
Hence,
\displaystyle \frac{d}{dx}\Big((\log x)^{\cos x}\Big)=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\,\log(\log x)\right]
Question 4: Exercise 5.5
4. Differentiate x^x-2^{\sin x} with respect to x.
Solution
Let
y=u-v\qquad ...(1)
where
u=x^x,\qquad v=2^{\sin x}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M-N)\ne \log M-\log N
Therefore, we differentiate u and v separately.
First, consider
u=x^x\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log(x^x)
=x\log x\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x\log x)
Using the Product Rule,
\displaystyle \frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=x\left(\frac{1}{x}\right)+\log x(1)=1+\log x\qquad ...(4)
Multiplying both sides of (4) by u, we get
\displaystyle \frac{du}{dx}=u(1+\log x)
Using equation (2),
\displaystyle \frac{du}{dx}=x^x(1+\log x)\qquad ...(5)
Now, consider
v=2^{\sin x}\qquad ...(6)
Taking logarithm on both sides, we get
\log v=\log\left(2^{\sin x}\right)
=(\sin x)\log 2\qquad ...(7)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (7) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=(\log 2)\cos x\qquad ...(8)
Multiplying both sides of (8) by v, we get
\displaystyle \frac{dv}{dx}=v(\log 2)\cos x
Using equation (6),
\displaystyle \frac{dv}{dx}=2^{\sin x}(\log 2)\cos x\qquad ...(9)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}
Using (5) and (9),
\displaystyle \frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}(\log 2)\cos x
Hence,
\displaystyle \frac{d}{dx}\left(x^x-2^{\sin x}\right)=x^x(1+\log x)-2^{\sin x}(\log 2)\cos x
Question 5: Logarithmic Differentiation 5.5
5. Differentiate (x+3)^2(x+4)^3(x+5)^4 with respect to x.
Solution
Let
y=(x+3)^2(x+4)^3(x+5)^4\qquad ...(1)
Taking logarithm on both sides, we get
\log y=\log\Big((x+3)^2(x+4)^3(x+5)^4\Big)
=\log(x+3)^2+\log(x+4)^3+\log(x+5)^4\qquad ...(2)
(Using \log(MNP)=\log M+\log N+\log P)
Applying the logarithmic property \log(M^n)=n\log M, equation (2) becomes
\log y=2\log(x+3)+3\log(x+4)+4\log(x+5)\qquad ...(3)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=2\frac{d}{dx}\big(\log(x+3)\big)+3\frac{d}{dx}\big(\log(x+4)\big)+4\frac{d}{dx}\big(\log(x+5)\big)
Using the Chain Rule,
\displaystyle \frac{d}{dx}\big(\log u\big)=\frac{1}{u}\frac{du}{dx}
Therefore,
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\qquad ...(4)
Multiplying both sides of (4) by y, we get
\displaystyle \frac{dy}{dx}=y\left(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right)\qquad ...(5)
Using equation (1) in (5),
We get, \displaystyle \frac{dy}{dx}=(x+3)^2(x+4)^3(x+5)^4\left(\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right)
\displaystyle =(x+3)(x+4)^2(x+5)^3\Big[2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)\Big]
\displaystyle =(x+3)(x+4)^2(x+5)^3\Big[(2x^2+18x+40)+(3x^2+24x+45)+(4x^2+28x+48)\Big]
i.e.
\displaystyle =(x+3)(x+4)^2(x+5)^3(9x^2+70x+133)
Hence,
\displaystyle \frac{d}{dx}\Big((x+3)^2(x+4)^3(x+5)^4\Big)=(x+3)(x+4)^2(x+5)^3(9x^2+70x+133)
Important: Before applying logarithmic differentiation, it is important to recognise the structure of a function. Identifying products, quotients, powers and variable exponents correctly often determines the most efficient method of differentiation.
Question 6: Differentiation by Using Logarithm
6. Differentiate \left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=\left(x+\frac{1}{x}\right)^x,\qquad v=x^{\left(1+\frac{1}{x}\right)}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=\left(x+\frac{1}{x}\right)^x\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log\left[\left(x+\frac{1}{x}\right)^x\right]
=x\log\left(x+\frac{1}{x}\right)\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left[x\log\left(x+\frac{1}{x}\right)\right]
Using the Product Rule,
\displaystyle \frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=\log\left(x+\frac{1}{x}\right)+x\cdot\frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}\qquad ...(4)
Multiplying both sides of (4) by u,
We get, \displaystyle \frac{du}{dx}=u\left[\log\left(x+\frac{1}{x}\right)+x\cdot\frac{1-\frac{1}{x^2}}{x+\frac{1}{x}}\right]
\displaystyle =u\left[\log\left(x+\frac{1}{x}\right)+\frac{x-\frac1x}{x+\frac1x}\right]
\displaystyle =u\left[\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]
Using equation (2),
\displaystyle \frac{du}{dx}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log\left(x+\frac{1}{x}\right)\right]\qquad ...(5)
Now, consider
v=x^{\left(1+\frac{1}{x}\right)}\qquad ...(6)
Taking logarithm on both sides, we get
\log v=\left(1+\frac{1}{x}\right)\log x\qquad ...(7)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (7) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=\left(1+\frac{1}{x}\right)\frac{1}{x}+\log x\left(-\frac{1}{x^2}\right)\qquad ...(8)
(Using the Product Rule)
Multiplying both sides of (8) by v, we get
\displaystyle \frac{dv}{dx}=v\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{\log x}{x^2}\right]
\displaystyle =v\left[\frac{x+1-\log x}{x^2}\right]
Using equation (6),
\displaystyle \frac{dv}{dx}=x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^2}\right)\qquad ...(9)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (5) and (9),
\displaystyle \frac{dy}{dx}=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log\left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^2}\right)
Hence,
\displaystyle \frac{d}{dx}\left[\left(x+\frac{1}{x}\right)^x+x^{\left(1+\frac{1}{x}\right)}\right]=\left(x+\frac{1}{x}\right)^x\left[\frac{x^2-1}{x^2+1}+\log\left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^2}\right)
Question 7: Logarithmic Differentiation 5.5
7. Differentiate (\log x)^x+x^{\log x} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=(\log x)^x,\qquad v=x^{\log x}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=(\log x)^x\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log\Big((\log x)^x\Big)
=x\log(\log x)\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\big(x\log(\log x)\big)
Using the Product Rule,
\displaystyle \frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=x\left(\frac{1}{x\log x}\right)+\log(\log x)(1)
\displaystyle =\frac{1}{\log x}+\log(\log x)\qquad ...(4)
Multiplying both sides of (4) by u, we get
\displaystyle \frac{du}{dx}=u\left(\frac{1}{\log x}+\log(\log x)\right)
Using equation (2),
\displaystyle \frac{du}{dx}=(\log x)^x\left(\frac{1}{\log x}+\log(\log x)\right)
Taking LCM inside the bracket, we get
\displaystyle \frac{du}{dx}=(\log x)^x\left(\frac{1+(\log x)\log(\log x)}{\log x}\right)
\displaystyle =(\log x)^{x-1}\Big[(\log x)\log(\log x)+1\Big]\qquad ...(5)
Now, consider
v=x^{\log x}\qquad ...(6)
Taking logarithm on both sides, we get
\log v=\log x\cdot\log x=(\log x)^2\qquad ...(7)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (7) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=2(\log x)\cdot\frac{1}{x}\qquad ...(8)
(Using the Chain Rule)
Multiplying both sides of (8) by v, we get
\displaystyle \frac{dv}{dx}=v\frac{2\log x}{x}
Using equation (6),
We get, \displaystyle \frac{dv}{dx}=x^{\log x}\frac{2\log x}{x}
\displaystyle =2(\log x)\frac{x^{\log x}}{x}
\displaystyle =2(\log x)\,x^{\log x-1}\qquad ...(9)
Differentiating both sides of equation (1) w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (5) and (9),
\displaystyle \frac{dy}{dx}=(\log x)^{x-1}\Big[(\log x)\log(\log x)+1\Big]+2(\log x)\,x^{\log x-1}
Hence,
\displaystyle \frac{d}{dx}\Big((\log x)^x+x^{\log x}\Big)=(\log x)^{x-1}\Big[(\log x)\log(\log x)+1\Big]+2(\log x)\,x^{\log x-1}
Exam Tip: Matching your final answer with the NCERT answer is a good practice because it helps in verifying your work and learning useful simplifications. However, during an examination you do not have access to the answer key. Therefore, focus on obtaining a correct and reasonably simplified answer using valid mathematical steps. There is no fixed rule about how far an answer must be simplified unless specifically required. Any algebraically equivalent form obtained through correct working is considered correct.
Now, let’s continue.
Question 8: Logarithmic Differentiation
8. Differentiate (\sin x)^x+\sin^{-1}\sqrt{x} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=(\sin x)^x,\qquad v=\sin^{-1}\sqrt{x}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=(\sin x)^x\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log\Big((\sin x)^x\Big)
=x\log(\sin x)\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\big(x\log(\sin x)\big)
Using the Product Rule,
\displaystyle \frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=x\left(\frac{\cos x}{\sin x}\right)+\log(\sin x)(1)
\displaystyle =x\cot x+\log(\sin x)\qquad ...(4)
Multiplying both sides of (4) by u, we get
\displaystyle \frac{du}{dx}=u\Big(x\cot x+\log(\sin x)\Big)
Using equation (2),
\displaystyle \frac{du}{dx}=(\sin x)^x\Big(x\cot x+\log(\sin x)\Big)\qquad ...(5)
Now, consider
v=\sin^{-1}\sqrt{x}\qquad ...(6)
Let
t=\sqrt{x}\qquad ...(7)
Then
v=\sin^{-1}t
Using the Chain Rule,
\displaystyle \frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}
Now,
\displaystyle \frac{dv}{dt}=\frac{1}{\sqrt{1-t^2}}
and
\displaystyle \frac{dt}{dx}=\frac{1}{2\sqrt{x}}
Therefore,
\displaystyle \frac{dv}{dx}=\frac{1}{\sqrt{1-t^2}}\cdot\frac{1}{2\sqrt{x}}
Using equation (7),
We get, \displaystyle \frac{dv}{dx}=\frac{1}{2\sqrt{x}\sqrt{1-x}}
\displaystyle =\frac{1}{2\sqrt{x(1-x)}}
\displaystyle =\frac{1}{2\sqrt{x-x^2}}\qquad ...(8)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (5) and (8),
\displaystyle \frac{dy}{dx}=(\sin x)^x\Big(x\cot x+\log(\sin x)\Big)+\frac{1}{2\sqrt{x-x^2}}
Hence,
\displaystyle \frac{d}{dx}\Big((\sin x)^x+\sin^{-1}\sqrt{x}\Big)=(\sin x)^x\Big(x\cot x+\log(\sin x)\Big)+\frac{1}{2\sqrt{x-x^2}}
Question 9: Logarithmic Differentiation 5.5
9. Differentiate x^{\sin x}+(\sin x)^{\cos x} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=x^{\sin x},\qquad v=(\sin x)^{\cos x}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=x^{\sin x}\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log(x^{\sin x})
=(\sin x)\log x\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\Big((\sin x)\log x\Big)
Using the Product Rule,
\displaystyle \frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=(\sin x)\left(\frac{1}{x}\right)+(\log x)(\cos x)
\displaystyle =\frac{\sin x}{x}+\cos x\log x\qquad ...(4)
Multiplying both sides of (4) by u, we get
\displaystyle \frac{du}{dx}=u\left(\frac{\sin x}{x}+\cos x\log x\right)
Using equation (2),
\displaystyle \frac{du}{dx}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\log x\right)\qquad ...(5)
Now, consider
v=(\sin x)^{\cos x}\qquad ...(6)
Taking logarithm on both sides, we get
\log v=\log\Big((\sin x)^{\cos x}\Big)
=(\cos x)\log(\sin x)\qquad ...(7)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (7) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=\cos x\cdot\frac{\cos x}{\sin x}+\log(\sin x)(-\sin x)
\displaystyle =\cos x\cot x-\sin x\log(\sin x)\qquad ...(8)
(Using the Product Rule)
Multiplying both sides of (8) by v, we get
\displaystyle \frac{dv}{dx}=v\Big(\cos x\cot x-\sin x\log(\sin x)\Big)
Using equation (6),
\displaystyle \frac{dv}{dx}=(\sin x)^{\cos x}\Big(\cos x\cot x-\sin x\log(\sin x)\Big)\qquad ...(9)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (5) and (9),
\displaystyle \frac{dy}{dx}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\log x\right)+(\sin x)^{\cos x}\Big(\cos x\cot x-\sin x\log(\sin x)\Big)
Hence,
\displaystyle \frac{d}{dx}\Big(x^{\sin x}+(\sin x)^{\cos x}\Big)=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\log x\right)+(\sin x)^{\cos x}\Big(\cos x\cot x-\sin x\log(\sin x)\Big)
Question 10: Logarithmic Differentiation
10. Differentiate x^{x\cos x}+\frac{x^2+1}{x^2-1} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=x^{x\cos x},\qquad v=\frac{x^2+1}{x^2-1}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=x^{x\cos x}\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log\Big(x^{x\cos x}\Big)
=x\cos x\log x\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x\cos x\log x)
Using the Product Rule for three factors,
\displaystyle \frac{1}{u}\frac{du}{dx}=\cos x\log x+x(-\sin x)\log x+x\cos x\left(\frac{1}{x}\right)
\displaystyle =\cos x\log x-x\sin x\log x+\cos x\qquad ...(4)
Rearranging the terms,
\displaystyle \frac{1}{u}\frac{du}{dx}=\cos x(1+\log x)-x\sin x\log x\qquad ...(5)
Multiplying both sides of (5) by u, we get
\displaystyle \frac{du}{dx}=u\Big[\cos x(1+\log x)-x\sin x\log x\Big]
Using equation (2),
\displaystyle \frac{du}{dx}=x^{x\cos x}\Big[\cos x(1+\log x)-x\sin x\log x\Big]\qquad ...(6)
Now, consider
v=\frac{x^2+1}{x^2-1}\qquad ...(7)
Using the Quotient Rule,
We get, \displaystyle \frac{dv}{dx}=\frac{(x^2-1)\frac{d}{dx}(x^2+1)-(x^2+1)\frac{d}{dx}(x^2-1)}{(x^2-1)^2}
\displaystyle =\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}
\displaystyle =\frac{2x\Big[(x^2-1)-(x^2+1)\Big]}{(x^2-1)^2}
or
\displaystyle =\frac{2x(-2)}{(x^2-1)^2}
\displaystyle =-\frac{4x}{(x^2-1)^2}\qquad ...(8)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (6) and (8),
\displaystyle \frac{dy}{dx}=x^{x\cos x}\Big[\cos x(1+\log x)-x\sin x\log x\Big]-\frac{4x}{(x^2-1)^2}
Hence,
\displaystyle \frac{d}{dx}\left(x^{x\cos x}+\frac{x^2+1}{x^2-1}\right)=x^{x\cos x}\Big[\cos x(1+\log x)-x\sin x\log x\Big]-\frac{4x}{(x^2-1)^2}
Choosing an appropriate method is often as important as performing the differentiation itself. Logarithmic differentiation provides an elegant approach for handling certain functions that are otherwise difficult to differentiate directly.
Question 11: Logarithmic Differentiation 5.5
11. Differentiate (x\cos x)^x+(x\sin x)^{\frac{1}{x}} with respect to x.
Solution
Let
y=u+v\qquad ...(1)
where
u=(x\cos x)^x,\qquad v=(x\sin x)^{\frac{1}{x}}
Common Mistake: We cannot apply logarithm directly to equation (1) because
\log(M+N)\ne \log M+\log N
Therefore, we differentiate u and v separately.
First, consider
u=(x\cos x)^x\qquad ...(2)
Taking logarithm on both sides, we get
\log u=\log\Big((x\cos x)^x\Big)
=x\log(x\cos x)\qquad ...(3)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (3) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\Big(x\log(x\cos x)\Big)
Using the Product Rule,
\displaystyle \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}\big(\log(x\cos x)\big)+\log(x\cos x)\frac{d}{dx}(x)
\displaystyle =x\left(\frac{1}{x\cos x}\cdot\frac{d}{dx}(x\cos x)\right)+\log(x\cos x)
Using the Product Rule on x\cos x,
\displaystyle \frac{d}{dx}(x\cos x)=\cos x-x\sin x
Therefore,
\displaystyle \frac{1}{u}\frac{du}{dx}=x\left(\frac{\cos x-x\sin x}{x\cos x}\right)+\log(x\cos x)
\displaystyle =1-x\tan x+\log(x\cos x)\qquad ...(4)
Multiplying both sides of (4) by u, we get
\displaystyle \frac{du}{dx}=u\Big[1-x\tan x+\log(x\cos x)\Big]
Using equation (2),
\displaystyle \frac{du}{dx}=(x\cos x)^x\Big[1-x\tan x+\log(x\cos x)\Big]\qquad ...(5)
Now, consider
v=(x\sin x)^{\frac{1}{x}}\qquad ...(6)
Taking logarithm on both sides, we get
\log v=\frac{1}{x}\log(x\sin x)\qquad ...(7)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (7) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{x}\log(x\sin x)\right)
Using the Product Rule,
\displaystyle \frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\frac{d}{dx}\big(\log(x\sin x)\big)+\log(x\sin x)\frac{d}{dx}\left(\frac{1}{x}\right)
\displaystyle =\frac{1}{x}\left(\frac{1}{x\sin x}\cdot\frac{d}{dx}(x\sin x)\right)-\frac{\log(x\sin x)}{x^2}
Using the Product Rule on x\sin x,
\displaystyle \frac{d}{dx}(x\sin x)=\sin x+x\cos x
Therefore,
We get, \displaystyle \frac{1}{v}\frac{dv}{dx}=\frac{1}{x}\left(\frac{\sin x+x\cos x}{x\sin x}\right)-\frac{\log(x\sin x)}{x^2}
\displaystyle =\frac{1+x\cot x}{x^2}-\frac{\log(x\sin x)}{x^2}
\displaystyle =\frac{x\cot x+1-\log(x\sin x)}{x^2}\qquad ...(8)
Multiplying both sides of (8) by v, we get
\displaystyle \frac{dv}{dx}=v\left[\frac{x\cot x+1-\log(x\sin x)}{x^2}\right]
Using equation (6),
\displaystyle \frac{dv}{dx}=(x\sin x)^{\frac{1}{x}}\left[\frac{x\cot x+1-\log(x\sin x)}{x^2}\right]\qquad ...(9)
Differentiating equation (1), we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
Using (5) and (9),
\displaystyle \frac{dy}{dx}=(x\cos x)^x\Big[1-x\tan x+\log(x\cos x)\Big]+(x\sin x)^{\frac{1}{x}}\left[\frac{x\cot x+1-\log(x\sin x)}{x^2}\right]
Hence,
\displaystyle \frac{d}{dx}\Big((x\cos x)^x+(x\sin x)^{\frac{1}{x}}\Big)=(x\cos x)^x\Big[1-x\tan x+\log(x\cos x)\Big]+(x\sin x)^{\frac{1}{x}}\left[\frac{x\cot x+1-\log(x\sin x)}{x^2}\right]
Question 12: Logarithmic Differentiation 5.5
12. Find \displaystyle \frac{dy}{dx} if x^y+y^x=a^b.
Solution
Given that
x^y+y^x=a^b\qquad ...(1)
Let
u=x^y,\qquad v=y^x
Then equation (1) becomes
u+v=a^b
Differentiating both sides w.r.t. x, we get
\displaystyle \frac{du}{dx}+\frac{dv}{dx}=0\qquad ...(2)
Now, consider
u=x^y\qquad ...(3)
Taking logarithm on both sides,
\log u=y\log x\qquad ...(4)
Differentiating equation (4), we get
\displaystyle \frac{1}{u}\frac{du}{dx}=y\frac{1}{x}+\log x\frac{dy}{dx}
\displaystyle \frac{du}{dx}=x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)\qquad ...(5)
Similarly,
v=y^x\qquad ...(6)
Taking logarithm on both sides,
\log v=x\log y\qquad ...(7)
Differentiating equation (7), we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}
\displaystyle \frac{dv}{dx}=y^x\left(\log y+\frac{x}{y}\frac{dy}{dx}\right)\qquad ...(8)
Substituting (5) and (8) into (2), we get
\displaystyle x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)+y^x\left(\log y+\frac{x}{y}\frac{dy}{dx}\right)=0
\displaystyle \left(x^y\log x+\frac{x}{y}y^x\right)\frac{dy}{dx}+\frac{yx^y}{x}+y^x\log y=0
or
\displaystyle \left(x^y\log x+xy^{x-1}\right)\frac{dy}{dx}= -\left(x^{y-1}y+y^x\log y\right)
Therefore,
\displaystyle \frac{dy}{dx}=-\frac{yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move on to the next question of Continuity and Differentiability 5.5.
Question 13: Differentiation 5.5
13. Find \displaystyle \frac{dy}{dx} if x^y=y^x.
Solution
Given that
x^y=y^x\qquad ...(1)
Taking logarithm on both sides, we get
\log(x^y)=\log(y^x)
y\log x=x\log y\qquad ...(2)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle y\frac{1}{x}+\log x\frac{dy}{dx}=\log y+x\frac{1}{y}\frac{dy}{dx}
\displaystyle \log x\frac{dy}{dx}-\frac{x}{y}\frac{dy}{dx}=\log y-\frac{y}{x}
or \displaystyle \left(\log x-\frac{x}{y}\right)\frac{dy}{dx}=\log y-\frac{y}{x}\qquad ...(3)
Therefore,
\displaystyle \frac{dy}{dx}=\frac{\log y-\frac{y}{x}}{\log x-\frac{x}{y}}
Multiplying the numerator and denominator by xy,
We get, \displaystyle \frac{dy}{dx}=\frac{xy\log y-y^2}{xy\log x-x^2}
\displaystyle =\frac{y(x\log y-y)}{x(y\log x-x)}
\displaystyle =\frac{y}{x}\cdot\frac{x\log y-y}{y\log x-x}
Multiplying the numerator and denominator of the fraction by -1, we get
\displaystyle \frac{dy}{dx}=\frac{y}{x}\left(\frac{y-x\log y}{x-y\log x}\right)
Hence,
\displaystyle \frac{dy}{dx}=\frac{y}{x}\left(\frac{y-x\log y}{x-y\log x}\right)
Question 14: Logarithmic Differentiation 5.5
14. Find \displaystyle \frac{dy}{dx} if (\cos x)^y=(\cos y)^x.
Solution
Given that
(\cos x)^y=(\cos y)^x\qquad ...(1)
Taking logarithm on both sides, we get
\log\Big((\cos x)^y\Big)=\log\Big((\cos y)^x\Big)
y\log(\cos x)=x\log(\cos y)\qquad ...(2)
(Using \log(M^n)=n\log M)
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle y\frac{d}{dx}\big(\log(\cos x)\big)+\log(\cos x)\frac{dy}{dx}=\log(\cos y)+x\frac{d}{dx}\big(\log(\cos y)\big)
Using the Chain Rule,
\displaystyle -y\tan x+\log(\cos x)\frac{dy}{dx}=\log(\cos y)-x\tan y\frac{dy}{dx}
Collecting the terms containing \frac{dy}{dx} on the left side, we get
\displaystyle \left[\log(\cos x)+x\tan y\right]\frac{dy}{dx}=\log(\cos y)+y\tan x\qquad ...(3)
Therefore,
\displaystyle \frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}
Hence,
\displaystyle \frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}
Question 15: Continuity and Differentiability 5.5
15. Find \displaystyle \frac{dy}{dx} if xy=e^{(x-y)}.
Solution
Given that
xy=e^{(x-y)}\qquad ...(1)
Taking logarithm on both sides, we get
\log(xy)=\log\left(e^{(x-y)}\right)
\log x+\log y=x-y\qquad ...(2)
(Using \log(MN)=\log M+\log N and \log(e^t)=t)
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}
Collecting the terms containing \frac{dy}{dx} on the right side, we get
\displaystyle \frac{1}{x}-1=-\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}
or \displaystyle \frac{1-x}{x}=-\left(1+\frac{1}{y}\right)\frac{dy}{dx}
or \displaystyle \frac{x-1}{x}=\frac{y+1}{y}\frac{dy}{dx}
Therefore,
\displaystyle \frac{dy}{dx}=\frac{x-1}{x}\cdot\frac{y}{y+1}
\displaystyle \frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}
Hence,
\displaystyle \frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}
Important: Before starting a differentiation problem, spend a few moments analysing the structure of the function. Choosing the right approach often reduces both the length of the solution and the chances of making mistakes.
Question 16: Logarithmic Differentiation 5.5
16. Find the derivative of the function f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8) and hence find f'(1).
Solution
Let
y=(1+x)(1+x^2)(1+x^4)(1+x^8)\qquad ...(1)
Taking logarithm on both sides, we get
\log y=\log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8)\qquad ...(2)
(Using \log(MNPQ)=\log M+\log N+\log P+\log Q)
Differentiating both sides of equation (2) w.r.t. x, we get
\displaystyle \frac1y\frac{dy}{dx}=\frac1{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\qquad ...(3)
Multiplying both sides of (3) by y, we get
\displaystyle \frac{dy}{dx}=y\left[\frac1{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]
Using equation (1),
\displaystyle f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)\left[\frac1{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]\qquad ...(4)
Putting x=1 in equation (4),
We get, \displaystyle f'(1)=(2)(2)(2)(2)\left[\frac12+\frac22+\frac42+\frac82\right]
\displaystyle =16(0.5+1+2+4)
\displaystyle =16\left(\frac{15}{2}\right)=120
Hence,
\displaystyle f'(1)=120
Question 17: Differentiation 5.5
17. Differentiate (x^2-5x+8)(x^3+7x+9) by using the Product Rule.
Method I: Using Product Rule
Let
u=x^2-5x+8,\qquad v=x^3+7x+9
Then
y=uv
Using the Product Rule,
\displaystyle \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}
Now,
\displaystyle \frac{du}{dx}=2x-5,\qquad \frac{dv}{dx}=3x^2+7
Therefore,
\displaystyle \frac{dy}{dx}=(x^2-5x+8)(3x^2+7)+(x^3+7x+9)(2x-5)
Expanding and simplifying,
\displaystyle \frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11
Method II: By Expanding First
(x^2-5x+8)(x^3+7x+9)=x^5-5x^4+15x^3-26x^2+11x+72
Differentiating term by term,
\displaystyle \frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11
Method III: By Logarithmic Differentiation
Let
y=(x^2-5x+8)(x^3+7x+9)
Taking logarithm on both sides,
\log y=\log(x^2-5x+8)+\log(x^3+7x+9)
Differentiating,
\displaystyle \frac1y\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}
Multiplying by y, we get
\displaystyle \frac{dy}{dx}=(x^3+7x+9)(2x-5)+(x^2-5x+8)(3x^2+7)
\displaystyle =5x^4-20x^3+45x^2-52x+11
Thus, all three methods give the same derivative.
Question 18: Logarithmic Differentiation 5.5
18. Show that
\displaystyle \frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}
where u,\;v and w are functions of x.
Method I: Repeated Application of Product Rule
Let
y=uvw=(uv)w\qquad ...(1)
Differentiating equation (1) w.r.t. x, we get
\displaystyle \frac{dy}{dx}=(uv)\frac{dw}{dx}+w\frac{d}{dx}(uv)\qquad ...(2)
(Using the Product Rule)
Again applying the Product Rule to uv,
\displaystyle \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\qquad ...(3)
Substituting (3) into (2), we get
\displaystyle \frac{dy}{dx}=uv\frac{dw}{dx}+w\left(u\frac{dv}{dx}+v\frac{du}{dx}\right)
\displaystyle \frac{dy}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}
Therefore,
\displaystyle \frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}
Method II: Logarithmic Differentiation
Let
y=uvw\qquad ...(4)
Taking logarithm on both sides, we get
\log y=\log u+\log v+\log w\qquad ...(5)
(Using \log(MNP)=\log M+\log N+\log P)
Differentiating both sides of equation (5) w.r.t. x, we get
\displaystyle \frac1y\frac{dy}{dx}=\frac1u\frac{du}{dx}+\frac1v\frac{dv}{dx}+\frac1w\frac{dw}{dx}\qquad ...(6)
Multiplying both sides of (6) by y=uvw, we get
\displaystyle \frac{dy}{dx}=uvw\left(\frac1u\frac{du}{dx}+\frac1v\frac{dv}{dx}+\frac1w\frac{dw}{dx}\right)
\displaystyle \frac{dy}{dx}=vw\frac{du}{dx}+uw\frac{dv}{dx}+uv\frac{dw}{dx}
Therefore,
\displaystyle \frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}
Hence proved.
Common Mistakes to Avoid
- Taking logarithm of a sum or difference: Remember that \log(M+N)\ne \log M+\log N and \log(M-N)\ne \log M-\log N. Logarithmic differentiation can be applied directly only when the expression is a product, quotient or power.
- Forgetting to take logarithm on both sides: If y=u(x)^{v(x)}, first write \log y=v(x)\log(u(x)). Applying ordinary differentiation directly is usually difficult.
- Not using logarithmic properties correctly: Always simplify expressions using identities such as \log(MN)=\log M+\log N, \log\left(\frac{M}{N}\right)=\log M-\log N and \log(M^n)=n\log M before differentiating.
- Forgetting implicit differentiation: After differentiating \log y, the result is \frac{1}{y}\frac{dy}{dx}, not simply \frac{1}{y}.
- Not multiplying back by y: After obtaining \frac{1}{y}\frac{dy}{dx}, multiply both sides by y and then substitute the original expression for y.
- Making errors while differentiating products: Expressions such as x\log x and x\log(\log x) require the Product Rule.
- Ignoring the Chain Rule inside logarithms: For example, \frac{d}{dx}\big(\log(\sin x)\big)=\frac{\cos x}{\sin x}=\cot x, not simply \frac{1}{\sin x}.
- Leaving the answer in a complicated form: After differentiation, simplify expressions wherever possible to match standard forms and NCERT answers.
- Confusing when logarithmic differentiation is useful: This method is especially useful for functions of the form u(x)^{v(x)} and for products or quotients involving many factors.
- Comparing only the appearance of answers: Two answers may look different but still be algebraically equivalent. Simplifying and matching NCERT answers is a good practice, but in examinations any mathematically equivalent form is acceptable.
Continue Learning
After completing Exercise 5.5 of Continuity and Differentiability, you should now be comfortable with logarithmic differentiation, differentiating functions of the form u(x)^{v(x)}, applying logarithmic properties to simplify complex expressions, and using logarithmic differentiation in both explicit and implicit functions.
To strengthen your understanding further, make sure that you revise:
- Meaning and purpose of logarithmic differentiation
- Situations where logarithmic differentiation is more convenient than ordinary differentiation
- Functions of the form u(x)^{v(x)}, where both the base and exponent are variable
- Taking logarithm on both sides and using implicit differentiation
- Important logarithmic properties such as \log(MN), \log\left(\frac{M}{N}\right) and \log(M^n)
- Using logarithmic properties to convert products into sums and quotients into differences
- Differentiating products and quotients involving many factors using logarithmic differentiation
- Combining logarithmic differentiation with the Chain Rule
- Combining logarithmic differentiation with the Product Rule and Quotient Rule
- Implicit differentiation of equations involving variable exponents
- Recognising mathematically equivalent forms of answers obtained after simplification
- Applications of logarithmic differentiation in simplifying lengthy differentiation problems
Explore More
Try differentiating a variety of functions involving variable bases and variable exponents using logarithmic differentiation. Practise taking logarithms on both sides, applying logarithmic properties correctly and then using implicit differentiation. Also solve questions involving products, quotients and powers with multiple factors, where logarithmic differentiation provides a shorter and more systematic approach than ordinary differentiation. Regular practice will help you identify the most suitable method and improve accuracy in solving complex differentiation problems.
All the best and keep learning 👍
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