Continuity and Differentiability 5.4 of NCERT Class 12 Maths Chapter 5 introduces the differentiation of exponential and logarithmic functions, which are among the most important functions in higher mathematics. In this exercise, we learn how to differentiate functions involving e^x, b^x, \ln x and \log_b x.
The number e, known as Euler’s number, plays a special role in calculus because the function e^x is the only exponential function whose derivative is equal to itself. Logarithmic functions are closely related to exponential functions, and understanding their properties makes differentiation much easier.
Before solving the questions, let us revise the important concepts, properties and derivative formulas of exponential and logarithmic functions used throughout Exercise 5.4.
Key Concepts
1. Exponential Function
Let b>1 be a real number. The function defined by
y=f(x)=b^x
is called an exponential function with base b.
The domain of an exponential function is all real numbers and its range is all positive real numbers.
- Domain: \mathbb{R}
- Range: (0,\infty)
Example: 2^x,\;5^x,\;10^x,\;e^x
2. Common and Natural Exponential Functions
An exponential function with base 10 is called a common exponential function.
y=10^x
An exponential function with base e is called a natural exponential function.
y=e^x
Here e\approx2.71828 is called Euler’s number.
The function e^x plays a special role in calculus because its derivative is equal to itself.
3. Logarithmic Function
Let b>1 be a real number. We say that logarithm of a to base b is x if
b^x=a
and write
x=\log_b a
The logarithmic function with base b is defined by
y=\log_b x
- Domain: (0,\infty)
- Range: \mathbb{R}
4. Common and Natural Logarithms
A logarithm with base 10 is called a common logarithm.
\log_{10}x
A logarithm with base e is called a natural logarithm and is denoted by
\ln x=\log_e x
The natural logarithm and natural exponential functions are inverse functions of each other.
5. Relationship Between Exponential and Logarithmic Functions
Exponential and logarithmic functions are inverse functions.
If
y=b^x
then
x=\log_b y
In particular,
e^{\ln x}=x
and
\ln(e^x)=x
6. Important Logarithmic Properties
- \log_b(MN)=\log_bM+\log_bN
- \log_b\left(\frac{M}{N}\right)=\log_bM-\log_bN
- \log_b(M^n)=n\log_bM
- \log_b1=0
- \log_bb=1
- \log_ba=\frac{\log_ca}{\log_cb}
Note: Throughout this exercise and elsewhere in NCERT Mathematics, the notation \log x is generally used for the natural logarithm. Thus,
\log x=\ln x=\log_e x
Therefore,
\displaystyle \frac{d}{dx}(\log x)=\frac{1}{x}
Properties of Logarithms
The following formulas are frequently used while solving differentiation problems. Memorising these standard results can make calculations faster and more accurate.
| 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) |
Derivative Formulas
| S.No. | Formula |
|---|---|
| 1. | \frac{d}{dx}(e^x)=e^x |
| 2. | \frac{d}{dx}(e^{u})=e^{u}\frac{du}{dx} |
| 3. | \frac{d}{dx}(a^x)=a^x\ln a |
| 4. | \frac{d}{dx}(a^{u})=a^{u}\ln a\cdot\frac{du}{dx} |
| 5. | \frac{d}{dx}(\ln x)=\frac{1}{x} |
| 6. | \frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx} |
| 7. | \frac{d}{dx}(\log_a x)=\frac{1}{x\ln a} |
Important: Exponential and logarithmic functions are inverses of each other. Their derivative formulas are among the most frequently used results in differential calculus and form the foundation for logarithmic differentiation in the next exercise.
Let us now solve all the NCERT questions step by step in this exercise 5.4 of Continuity and Differentiability.
Question 1: Continuity and Differentiability 5.4
1. Differentiate \displaystyle \frac{e^x}{\sin x} with respect to x.
Solution
Let
\displaystyle y=\frac{e^x}{\sin x}
This is a quotient of two functions. Therefore, we use the Quotient Rule:
\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}
Here,
u=e^x,\qquad v=\sin x
Now,
\displaystyle \frac{du}{dx}=e^x
and
\displaystyle \frac{dv}{dx}=\cos x
Applying the Quotient Rule,
\displaystyle \frac{dy}{dx}=\frac{\sin x\cdot e^x-e^x\cdot\cos x}{\sin^2 x}
Taking e^x common from the numerator,
\displaystyle \frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{\sin^2 x}
Hence,
\displaystyle \frac{d}{dx}\left(\frac{e^x}{\sin x}\right)=\frac{e^x(\sin x-\cos x)}{\sin^2 x}
Alternative Direct Method:
Using the Quotient Rule directly,
\displaystyle \frac{d}{dx}\left(\frac{e^x}{\sin x}\right)=\frac{\sin x\cdot\frac{d}{dx}(e^x)-e^x\cdot\frac{d}{dx}(\sin x)}{\sin^2 x}
\displaystyle =\frac{\sin x\cdot e^x-e^x\cos x}{\sin^2 x}
i.e.
\displaystyle =\frac{e^x(\sin x-\cos x)}{\sin^2 x}
Question 2: Differentiability 5.4
2. Differentiate e^{\sin^{-1}x} with respect to x.
Solution
Let
y=e^{\sin^{-1}x}
This is a composite function of the form
y=e^u
where
u=\sin^{-1}x
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=e^u
and
\displaystyle \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}
Therefore,
\displaystyle \frac{dy}{dx}=e^u\cdot\frac{1}{\sqrt{1-x^2}}
Substituting u=\sin^{-1}x, we get
\displaystyle \frac{dy}{dx}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}
Hence,
\displaystyle \frac{d}{dx}\left(e^{\sin^{-1}x}\right)=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}
You may already be following Maths Better for NCERT Solutions for the topics like Matrices, Determinants, Relations and Functions and Inverse Trigonometric Functions. Likewise this exercise of Continuity and Differentiability 5.4 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 5.4.
Question 3: Continuity and Differentiability 5.4
3. Differentiate e^{x^3} with respect to x.
Solution
Let
y=e^{x^3}
This is a composite function of the form
y=e^u
where
u=x^3
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=e^u
and
\displaystyle \frac{du}{dx}=3x^2
Therefore,
\displaystyle \frac{dy}{dx}=e^u\cdot3x^2
Substituting u=x^3, we get
\displaystyle \frac{dy}{dx}=3x^2e^{x^3}
Hence,
\displaystyle \frac{d}{dx}\left(e^{x^3}\right)=3x^2e^{x^3}
Question 4: Exercise 5.4
4. Differentiate \sin\left(\tan^{-1}e^{-x}\right) with respect to x.
Solution
Let
u=\tan^{-1}e^{-x}\qquad ...(1)
Then
y=\sin u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
i.e.
\displaystyle \frac{dy}{dx}=\cos u\cdot\frac{du}{dx}\qquad ...(2)
Now, differentiating equation (1),
We get, \displaystyle \frac{du}{dx}=\frac{1}{1+(e^{-x})^2}\cdot\frac{d}{dx}(e^{-x})
\displaystyle =\frac{1}{1+e^{-2x}}\cdot(-e^{-x})
\displaystyle =-\frac{e^{-x}}{1+e^{-2x}}\qquad ...(3)
Substituting (3) into (2), we get
\displaystyle \frac{dy}{dx}=-\frac{e^{-x}\cos u}{1+e^{-2x}}
Using equation (1),
\displaystyle \frac{dy}{dx}=-\frac{e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}
Hence,
\boxed{\displaystyle \frac{d}{dx}\left[\sin\left(\tan^{-1}e^{-x}\right)\right]=-\frac{e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}}
Question 5: Differentiability 5.4
5. Differentiate \log(\cos e^x) with respect to x.
Solution
Let
u=\cos e^x\qquad ...(1)
Then
y=\log u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}\qquad ...(2)
Now, differentiating equation (1),
We get, \displaystyle \frac{du}{dx}=-\sin(e^x)\cdot\frac{d}{dx}(e^x)
\displaystyle =-\sin(e^x)\cdot e^x
\displaystyle =-e^x\sin(e^x)\qquad ...(3)
Substituting (1) and (3) into (2), we get
\displaystyle \frac{dy}{dx}=\frac{-e^x\sin(e^x)}{\cos(e^x)}
Using \tan\theta=\frac{\sin\theta}{\cos\theta}, we get
\displaystyle \frac{dy}{dx}=-e^x\tan(e^x)
Hence,
\boxed{\displaystyle \frac{d}{dx}\big(\log(\cos e^x)\big)=-e^x\tan(e^x)}
Important: Exponential and logarithmic functions form an important class of functions in calculus. Since these functions frequently occur in mathematical models and applications, it is important to be familiar with their properties, standard derivatives and their use in composite functions.
Question 6: Continuity and Differentiability 5.4
6. Differentiate e^x+e^{x^2}+\cdots+e^{x^5} with respect to x.
Solution
Let
y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}\qquad ...(1)
Differentiating equation (1) with respect to x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(e^x)+\frac{d}{dx}(e^{x^2})+\frac{d}{dx}(e^{x^3})+\frac{d}{dx}(e^{x^4})+\frac{d}{dx}(e^{x^5})\qquad ...(2)
Using the formula
\displaystyle \frac{d}{dx}(e^u)=e^u\frac{du}{dx}\qquad ...(3)
Equation (2) becomes
\displaystyle \frac{dy}{dx}=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}
Hence,
\displaystyle \frac{d}{dx}\left(e^x+e^{x^2}+\cdots+e^{x^5}\right)=
\boxed{\displaystyle e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}}
Question 7: Differentiability 5.4
7. Differentiate \sqrt{e^{\sqrt{x}}}, x>0 with respect tox.
Solution
Let
y=\sqrt{e^{\sqrt{x}}}\qquad ...(1)
Using the law of exponents, equation (1) becomes
y=\left(e^{\sqrt{x}}\right)^{1/2}=e^{\frac{\sqrt{x}}{2}}\qquad ...(2)
Let
u=\frac{\sqrt{x}}{2}\qquad ...(3)
Then equation (2) can be written as
y=e^u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\qquad ...(4)
Now,
\displaystyle \frac{dy}{du}=e^u
and differentiating equation (3),
We get, \displaystyle \frac{du}{dx}=\frac{1}{2}\cdot\frac{d}{dx}(\sqrt{x})
\displaystyle =\frac{1}{2}\cdot\frac{1}{2\sqrt{x}}
\displaystyle =\frac{1}{4\sqrt{x}}\qquad ...(5)
Substituting in (4), we get
\displaystyle \frac{dy}{dx}=e^u\cdot\frac{1}{4\sqrt{x}}
Using (3),
\displaystyle \frac{dy}{dx}=\frac{e^{\frac{\sqrt{x}}{2}}}{4\sqrt{x}}
Hence,
\boxed{\displaystyle \frac{d}{dx}\left(\sqrt{e^{\sqrt{x}}}\right)=\frac{e^{\frac{\sqrt{x}}{2}}}{4\sqrt{x}}}
Note: The answer can also be written in the form given by NCERT.
Since
\displaystyle \sqrt{xe^{\sqrt{x}}}=\sqrt{x}\cdot e^{\frac{\sqrt{x}}{2}}\qquad ...(6)
therefore,
\displaystyle \frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}}=\frac{e^{\sqrt{x}}}{4\sqrt{x}\,e^{\frac{\sqrt{x}}{2}}}
\displaystyle =\frac{e^{\frac{\sqrt{x}}{2}}}{4\sqrt{x}}
Hence, both forms of the answer are equivalent.
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 continue.
Question 8: Continuity and Differentiability 5.4
8. Differentiate \log(\log x),\;x>1 with respect to x.
Solution
Let
u=\log x\qquad ...(1)
Then
y=\log u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}\qquad ...(2)
Now, differentiating equation (1), we get
\displaystyle \frac{du}{dx}=\frac{1}{x}\qquad ...(3)
Substituting (1) and (3) into (2), we get
\displaystyle \frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}
\displaystyle \frac{dy}{dx}=\frac{1}{x\log x}
Hence,
\boxed{\displaystyle \frac{d}{dx}\big(\log(\log x)\big)=\frac{1}{x\log x}}
The functions e^x and \log x occupy a special place in mathematics because they are inverse functions of each other. A strong understanding of their properties and derivatives forms the foundation for many advanced topics in calculus.
Question 9: Continuity and Differentiability Ex. 5.4
9. Differentiate \displaystyle \frac{\cos x}{\log x},\;x>0 with respect to x.
Solution
Let
\displaystyle u=\cos x,\qquad v=\log x\qquad ...(1)
Then
\displaystyle y=\frac{u}{v}
Using the Quotient Rule,
\displaystyle \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\qquad ...(2)
Now, differentiating equation (1), we get
\displaystyle \frac{du}{dx}=-\sin x,\qquad \frac{dv}{dx}=\frac{1}{x}\qquad ...(3)
Substituting (1) and (3) into (2), we get
\displaystyle \frac{dy}{dx}=\frac{\log x(-\sin x)-\cos x\left(\frac{1}{x}\right)}{(\log x)^2}
\displaystyle =\frac{-\log x\sin x-\frac{\cos x}{x}}{(\log x)^2}
Multiplying the numerator and denominator by x, we get
\displaystyle \frac{dy}{dx}=\frac{-x\log x\sin x-\cos x}{x(\log x)^2}
Hence,
\boxed{\displaystyle \frac{d}{dx}\left(\frac{\cos x}{\log x}\right)=\frac{-x\log x\sin x-\cos x}{x(\log x)^2}}
Question 10: Continuity and Differentiability 5.4
10. Differentiate \cos(\log x+e^x),\;x>0 with respect to x.
Solution
Let
u=\log x+e^x\qquad ...(1)
Then
y=\cos u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=-\sin u\cdot\frac{du}{dx}\qquad ...(2)
Now, differentiating equation (1), we get
\displaystyle \frac{du}{dx}=\frac{d}{dx}(\log x)+\frac{d}{dx}(e^x)
\displaystyle =\frac{1}{x}+e^x\qquad ...(3)
Substituting (1) and (3) into (2), we get
\displaystyle \frac{dy}{dx}=-\sin(\log x+e^x)\left(\frac{1}{x}+e^x\right)
Hence,
\boxed{\displaystyle \frac{d}{dx}\big(\cos(\log x+e^x)\big)=-\left(\frac{1}{x}+e^x\right)\sin(\log x+e^x)}
Common Mistakes to Avoid
- Forgetting the Chain Rule: Many functions in this exercise involve nested functions such as e^{x^2}, \log(\log x) and \cos(\log x+e^x). Always multiply by the derivative of the inner function.
- Differentiating e^u incorrectly: Students often write \frac{d}{dx}(e^u)=e^u. The correct formula is \frac{d}{dx}(e^u)=e^u\frac{du}{dx}.
- Forgetting that \log x denotes the natural logarithm: In NCERT, \log x is generally used for \ln x. Therefore, \frac{d}{dx}(\log x)=\frac{1}{x}.
- Differentiating logarithmic functions incorrectly: For \log(u), the derivative is not simply \frac{1}{u}. The correct formula is \frac{d}{dx}(\log u)=\frac{1}{u}\frac{du}{dx}.
- Ignoring the derivative of the exponent: While differentiating e^{x^3}, students often write only e^{x^3}. The correct derivative is 3x^2e^{x^3}.
- Applying inverse trigonometric derivatives incorrectly: Remember that \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} and \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}. These formulas are frequently used together with the Chain Rule.
- Making sign errors with negative exponents: For example, \frac{d}{dx}(e^{-x})=-e^{-x}. The negative sign is often missed.
- Using the Quotient Rule incorrectly: In expressions such as \frac{e^x}{\sin x} or \frac{\cos x}{\log x}, remember the formula \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}.
- Not simplifying the final answer: Expressions such as \frac{\sin\theta}{\cos\theta} should be simplified to \tan\theta whenever possible.
- Confusing exponential and logarithmic functions: Remember that e^x and \ln x are inverse functions. Useful identities are
Continue Learning
After completing Exercise 5.4 of Continuity and Differentiability, you should now be comfortable with differentiating exponential and logarithmic functions, applying the Chain Rule to composite functions involving e^x and \log x, and using the standard derivatives of exponential and logarithmic functions in a variety of problems.
To strengthen your understanding further, make sure that you revise:
- Definition of exponential functions of the form y=b^x, where b>1
- Domain and range of exponential functions
- Common exponential functions and natural exponential functions
- The special number e (Euler’s number) and its importance in calculus
- Definition of logarithms and the relationship between logarithmic and exponential functions
- Common logarithms and natural logarithms
- Basic properties of logarithms and the change of base formula
- Standard derivatives of e^x and \log x
- Derivative of e^{u} using the Chain Rule
- Derivative of \log(u) using the Chain Rule
- Differentiating composite functions involving exponential, logarithmic and trigonometric functions
- Applying Product Rule, Quotient Rule and Chain Rule together in a single problem
Explore More
Try differentiating a variety of exponential and logarithmic functions on your own. Practise applying the Chain Rule to functions involving e^x, e^{u}, \log x and \log(u). Also solve questions that combine exponential, logarithmic and trigonometric functions using the Product Rule and Quotient Rule. Regular practice will help you develop a systematic approach and improve accuracy in solving differentiation problems.
All the best and keep learning 👍
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