Differentiability 5.2 NCERT SolutionsThese NCERT Solutions for Differentiability 5.2 focus on the concept of differentiability which is an important concept in Calculus. The exercise examines the existence of derivatives, introduces important differentiation rules and develops techniques for finding derivatives of more complex functions.
A function f(x) is said to be differentiable at x = c if its derivative exists at that point, that is, if
\displaystyle f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}
exists and has the same value from both the left and right sides. In other words, the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) must be equal.
Exercise 5.2 also introduces important differentiation rules such as the sum rule, difference rule, product rule, quotient rule and constant multiple rule. You will further learn that every differentiable function is continuous, but a continuous function need not be differentiable. We shall also learn to find the derivatives of composite functions using the Chain Rule, which is a powerful technique for differentiating functions made up of other functions.
Key Concepts
1. Differentiability
A function is said to be differentiable at a point if its derivative exists at that point.
The derivative function is defined as:
\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
The derivative at x=c is obtained by substituting x=c in the above definition:
\displaystyle f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}
Example:
The function f(x)=x^2 is differentiable for all real values of x.
2. Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD)
To check differentiability at x=c, we compare the left-hand derivative and right-hand derivative.
Right-Hand Derivative (RHD) at x=c:
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}
Left-Hand Derivative (LHD)at x=c:
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(c-h)-f(c)}{-h}
A function is differentiable at x=c if LHD and RHD both exist and are equal.
3. Differentiability and Continuity
Every differentiable function is continuous, but every continuous function is not necessarily differentiable.
If a function is differentiable at a point, then it must be continuous at that point.
Example:
The function f(x)=|x| is continuous at x=0 but not differentiable there.
4. Basic Rules of Differentiation
The following rules simplify the process of finding derivatives.
- Sum Rule: \frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx}
- Difference Rule: \frac{d}{dx}(u-v)=\frac{du}{dx}-\frac{dv}{dx}
- Constant Multiple Rule: \frac{d}{dx}(cu)=c\frac{du}{dx}
- Product Rule: \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
- Quotient Rule: \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}
A useful way to remember the Quotient Rule is:
\displaystyle \frac{d}{dx}\left(\frac{N}{D}\right)=\frac{DN'-ND'}{D^2}
5. Derivative of Composite Functions
A composite function is formed when one function is applied to the output of another function.
It is generally written as:
y=f(g(x))
Here, g(x) is the inner function and f is the outer function.
Its derivative is given by:
\displaystyle \frac{d}{dx}\big(f(g(x))\big)=f'(g(x))\cdot g'(x)
6. Chain Rule
The Chain Rule is used to differentiate composite functions.
If y=f(u) and u=g(x), then
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Example:
For y=(2x+1)^3,
\frac{dy}{dx}=3(2x+1)^2\cdot 2=6(2x+1)^2
Important Differentiation Rules and Standard Derivatives
The following formulas are frequently used while solving differentiation problems. Memorising these standard results can make calculations faster and more accurate.
| S.No. | Rule / Formula |
|---|---|
| 1. | \displaystyle \frac{d}{dx}(c)=0 |
| 2. | \displaystyle \frac{d}{dx}(x)=1 |
| 3. | \displaystyle \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2} |
| 4. | \displaystyle \frac{d}{dx}(x^n)=nx^{n-1} |
| 5. | \displaystyle \frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2\sqrt{x}} |
| 6. | \displaystyle \frac{d}{dx}(cu)=c\frac{du}{dx} |
| 7. | \displaystyle \frac{d}{dx}(u+v)=\frac{du}{dx}+\frac{dv}{dx} |
| 8. | \displaystyle \frac{d}{dx}(u-v)=\frac{du}{dx}-\frac{dv}{dx} |
| 9. | \displaystyle \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx} |
| 10. | \displaystyle \frac{d}{dx}\left(\frac{N}{D}\right)=\frac{DN'-ND'}{D^2} |
| 11. | \displaystyle \frac{d}{dx}(\sin x)=\cos x |
| 12. | \displaystyle \frac{d}{dx}(\cos x)=-\sin x |
| 13. | \displaystyle \frac{d}{dx}(\tan x)=\sec^2x |
| 14. | \displaystyle \frac{d}{dx}(\cot x)=-\cosec^2x |
| 15. | \displaystyle \frac{d}{dx}(\sec x)=\sec x\tan x |
| 16. | \displaystyle \frac{d}{dx}(\cosec x)=-\cosec x\cot x |
| 17. | \displaystyle \frac{d}{dx}(e^x)=e^x |
| 18. | \displaystyle \frac{d}{dx}(a^x)=a^x\ln a |
| 19. | \displaystyle \frac{d}{dx}(\ln x)=\frac{1}{x} |
| 20. | \displaystyle \frac{d}{dx}(\log_a x)=\frac{1}{x\ln a} |
| 21. | \displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} |
Important: Formula 4 is the Power Rule, where n\in\mathbb{R}. Formula 9 is the Product Rule, Formula 10 is the Quotient Rule and Formula 21 is the Chain Rule.
Let us now solve all the NCERT questions step by step in this exercise 5.2 of Continuity and Differentiability.
Question 1: Differentiability 5.2 – Chain Rule
1. Differentiate \sin(x^2+5) with respect to x.
Solution
Let
y=\sin(x^2+5)
This is a composite function of the form
y=\sin(u)
where
u=x^2+5
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=\cos u
and
\displaystyle \frac{du}{dx}=2x
Therefore,
\displaystyle \frac{dy}{dx}=\cos u\cdot 2x
Substituting u=x^2+5, we get
\displaystyle \frac{dy}{dx}=2x\cos(x^2+5)
Hence,
\displaystyle \frac{d}{dx}\big(\sin(x^2+5)\big)=2x\cos(x^2+5)
Alternative Method:
Once you are comfortable with the Chain Rule, the differentiation can be done directly as follows:
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\big(\sin(x^2+5)\big)
\displaystyle =\cos(x^2+5)\cdot\frac{d}{dx}(x^2+5)
(Applying Chain Rule from outer to inner functions)
\displaystyle =\cos(x^2+5)\cdot 2x
\displaystyle =2x\cos(x^2+5)
Question 2: Differentiability 5.2
2. Differentiate \cos(\sin x) with respect to x.
Solution
Let
y=\cos(\sin x)
This is a composite function of the form
y=\cos u
where
u=\sin x
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=-\sin u
and
\displaystyle \frac{du}{dx}=\cos x
Therefore,
\displaystyle \frac{dy}{dx}=(-\sin u)(\cos x)
Substituting u=\sin x, we get
\displaystyle \frac{dy}{dx}=-\sin(\sin x)\cos x
Hence,
\displaystyle \frac{d}{dx}\big(\cos(\sin x)\big)=-\sin(\sin x)\cos x
Alternative Method:
Once you are comfortable with the Chain Rule, the differentiation can be done directly as follows:
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\big(\cos(\sin x)\big)
\displaystyle =-\sin(\sin x)\cdot\frac{d}{dx}(\sin x)
(Applying Chain Rule from outer to inner functions)
\displaystyle =-\sin(\sin x)\cdot\cos x
\displaystyle =-\sin(\sin x)\cos x
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 Differentiability 5.2 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 5.2.
Question 3: Continuity and Differentiability
3. Differentiate \sin(ax+b) with respect to x.
Solution
Let
y=\sin(ax+b)
This is a composite function of the form
y=\sin u
where
u=ax+b
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=\cos u
and
\displaystyle \frac{du}{dx}=a
Therefore,
\displaystyle \frac{dy}{dx}=a\cos u
Substituting u=ax+b, we get
\displaystyle \frac{dy}{dx}=a\cos(ax+b)
Hence,
\displaystyle \frac{d}{dx}\big(\sin(ax+b)\big)=a\cos(ax+b)
Alternative Method:
Once you are comfortable with the Chain Rule, the differentiation can be done directly as follows:
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\big(\sin(ax+b)\big)
\displaystyle =\cos(ax+b)\cdot\frac{d}{dx}(ax+b)
(Applying Chain Rule from outer to inner functions)
\displaystyle =\cos(ax+b)\cdot a
\displaystyle =a\cos(ax+b)
Question 4: Exercise 5.2
4. Differentiate \sec(\tan\sqrt{x}) with respect to x.
Solution
Let
y=\sec(\tan\sqrt{x})
Put
u=\tan\sqrt{x}
Then
y=\sec u
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=\sec u\tan u \qquad ...(1)
Also,
u=\tan\sqrt{x}
Therefore,
\displaystyle \frac{du}{dx}=\sec^2(\sqrt{x})\cdot\frac{d}{dx}(\sqrt{x}) \qquad ...(2)
Since
\displaystyle \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}} \qquad ...(3)
Substituting (3) into (2), we get
\displaystyle \frac{du}{dx}=\frac{\sec^2(\sqrt{x})}{2\sqrt{x}} \qquad ...(4)
Using (1) and (4) in
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
we get
\displaystyle \frac{dy}{dx}=\sec u\tan u\cdot\frac{\sec^2(\sqrt{x})}{2\sqrt{x}}
Substituting u=\tan\sqrt{x}, we obtain
\displaystyle \frac{dy}{dx}=\frac{\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\,\sec^2(\sqrt{x})}{2\sqrt{x}}
Hence,
\displaystyle \frac{d}{dx}\Big(\sec(\tan\sqrt{x})\Big)=\frac{\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\,\sec^2(\sqrt{x})}{2\sqrt{x}}
Alternative Method:
Once you are comfortable with the Chain Rule, the differentiation can be done directly as follows:
We get, \displaystyle \frac{dy}{dx}=\frac{d}{dx}\Big(\sec(\tan\sqrt{x})\Big)
\displaystyle =\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\cdot\frac{d}{dx}(\tan\sqrt{x})
\displaystyle =\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\cdot\sec^2(\sqrt{x})\cdot\frac{d}{dx}(\sqrt{x})
(Applying Chain Rule from outer to inner functions)
\displaystyle =\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\cdot\sec^2(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}
\displaystyle =\frac{\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\,\sec^2(\sqrt{x})}{2\sqrt{x}}
Question 5: Differentiability 5.2
5. Differentiate \displaystyle \frac{\sin(ax+b)}{\cos(cx+d)} with respect to x.
Solution
Let
\displaystyle y=\frac{\sin(ax+b)}{\cos(cx+d)}
Using the Quotient Rule,
\displaystyle \frac{dy}{dx}=\frac{D\,\frac{dN}{dx}-N\,\frac{dD}{dx}}{D^2}
where
N=\sin(ax+b)\quad\text{and}\quad D=\cos(cx+d)
Now,
\displaystyle \frac{dN}{dx}=a\cos(ax+b)\qquad ...(1)
Also,
\displaystyle \frac{dD}{dx}=-c\sin(cx+d)\qquad ...(2)
Using (1) and (2) in the Quotient Rule, we get
\displaystyle \frac{dy}{dx}=\frac{\cos(cx+d)\cdot a\cos(ax+b)-\sin(ax+b)\cdot\big(-c\sin(cx+d)\big)}{\cos^2(cx+d)}
Therefore,
\displaystyle \frac{dy}{dx}=\frac{a\cos(ax+b)\cos(cx+d)+c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}
Hence,
\displaystyle \frac{d}{dx}\left(\frac{\sin(ax+b)}{\cos(cx+d)}\right)=\frac{a\cos(ax+b)\cos(cx+d)+c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}
Important: The Chain Rule is one of the most important concepts in Differential Calculus. Whenever a function is formed by combining two or more functions, first identify the outer function and the inner function. Differentiate the outer function first and then multiply by the derivative of the inner function. For nested functions, the process is repeated from the outermost function to the innermost function.
Question 6: Differentiation by Chain Rule
6. Differentiate \cos(x^3)\cdot\sin^2(x^5) with respect to x.
Solution
Let
y=\cos(x^3)\cdot\sin^2(x^5)
This is a product of two functions. Therefore, we use the Product Rule:
\displaystyle \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
where
u=\cos(x^3)\quad\text{and}\quad v=\sin^2(x^5)
Now,
u=\cos(x^3)
Using the Chain Rule,
\displaystyle \frac{du}{dx}=-\sin(x^3)\cdot\frac{d}{dx}(x^3)
\displaystyle \frac{du}{dx}=-3x^2\sin(x^3)\qquad ...(1)
Also,
v=\sin^2(x^5)
Using the Chain Rule,
\displaystyle \frac{dv}{dx}=2\sin(x^5)\cdot\frac{d}{dx}\big(\sin(x^5)\big)
Since
\displaystyle \frac{d}{dx}\big(\sin(x^5)\big)=\cos(x^5)\cdot\frac{d}{dx}(x^5)
\displaystyle =5x^4\cos(x^5)\qquad ...(2)
Using (2), we get
\displaystyle \frac{dv}{dx}=10x^4\sin(x^5)\cos(x^5)\qquad ...(3)
Using (1) and (3) in the Product Rule, we get
\displaystyle \frac{dy}{dx}=\cos(x^3)\Big(10x^4\sin(x^5)\cos(x^5)\Big)+\sin^2(x^5)\Big(-3x^2\sin(x^3)\Big)
Therefore,
\displaystyle \frac{dy}{dx}=10x^4\cos(x^3)\sin(x^5)\cos(x^5)-3x^2\sin(x^3)\sin^2(x^5)
Hence,
\displaystyle \frac{d}{dx}\Big(\cos(x^3)\sin^2(x^5)\Big)=10x^4\cos(x^3)\sin(x^5)\cos(x^5)-3x^2\sin(x^3)\sin^2(x^5)
Question 7: Differentiability 5.2
7. Differentiate 2\sqrt{\cot(x^2)} with respect to x.
Solution
Let
y=2\sqrt{\cot(x^2)}
Using the Constant Multiple Rule,
\displaystyle \frac{dy}{dx}=2\frac{d}{dx}\Big(\sqrt{\cot(x^2)}\Big)
Now,
\displaystyle \sqrt{\cot(x^2)}=\big(\cot(x^2)\big)^{1/2}
Using the Chain Rule,
\displaystyle \frac{d}{dx}\Big(\sqrt{\cot(x^2)}\Big)=\frac{1}{2\sqrt{\cot(x^2)}}\cdot\frac{d}{dx}\big(\cot(x^2)\big)\qquad ...(1)
Also,
\displaystyle \frac{d}{dx}\big(\cot(x^2)\big)=-\csc^2(x^2)\cdot\frac{d}{dx}(x^2)
\displaystyle =-\csc^2(x^2)\cdot 2x\qquad ...(2)
Substituting (2) into (1), we get
\displaystyle \frac{d}{dx}\Big(\sqrt{\cot(x^2)}\Big)=\frac{1}{2\sqrt{\cot(x^2)}}\Big(-2x\,\csc^2(x^2)\Big)
\displaystyle =-\frac{x\,\csc^2(x^2)}{\sqrt{\cot(x^2)}}\qquad ...(3)
Using (3), we obtain
\displaystyle \frac{dy}{dx}=2\left(-\frac{x\,\csc^2(x^2)}{\sqrt{\cot(x^2)}}\right)
\displaystyle =-\frac{2x\,\csc^2(x^2)}{\sqrt{\cot(x^2)}}
Hence,
\displaystyle \frac{d}{dx}\Big(2\sqrt{\cot(x^2)}\Big)=-\frac{2x\,\csc^2(x^2)}{\sqrt{\cot(x^2)}}
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 5.2
8. Differentiate \cos(\sqrt{x}) with respect to x.
Solution
Let
y=\cos(\sqrt{x})
This is a composite function of the form
y=\cos u
where
u=\sqrt{x}
Using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
Now,
\displaystyle \frac{dy}{du}=-\sin u \qquad ...(1)
Also,
\displaystyle \frac{du}{dx}=\frac{d}{dx}(\sqrt{x})
\displaystyle =\frac{1}{2\sqrt{x}} \qquad ...(2)
Using (1) and (2) in the Chain Rule, we get
\displaystyle \frac{dy}{dx}=(-\sin u)\left(\frac{1}{2\sqrt{x}}\right)
\displaystyle =-\frac{\sin u}{2\sqrt{x}}
Substituting u=\sqrt{x}, we obtain
\displaystyle \frac{dy}{dx}=-\frac{\sin(\sqrt{x})}{2\sqrt{x}}
Hence,
\displaystyle \frac{d}{dx}\Big(\cos(\sqrt{x})\Big)=-\frac{\sin(\sqrt{x})}{2\sqrt{x}}
Functions such as modulus and greatest integer functions provide important examples of non-differentiability. Practising such questions helps in understanding how unequal or non-existent Left-Hand and Right-Hand Derivatives lead to failure of differentiability.
The following two questions from Exercise on Differentiability 5.2 are important from the examination point of view and should be practised carefully.
Question 9: Modulus Function is Non-Differentiable?
9. Prove that the function f(x)=|x-1|,\ x\in\mathbb{R} is not differentiable at x=1.
Solution
To prove that f(x)=|x-1| is not differentiable at x=1, we compare its Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at x=1.
Right-Hand Derivative (RHD)
Using the definition,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}
Since
f(1+h)=|(1+h)-1|=|h|
and for h>0, |h|=h, we have
f(1+h)=h
Also,
f(1)=|1-1|=0
Therefore,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{h-0}{h}
\displaystyle =\lim_{h\to0}1
i.e. RHD \displaystyle =1 \qquad ...(1)
Left-Hand Derivative (LHD)
Using the definition,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}
Since
f(1-h)=|(1-h)-1|=|-h|
and for h>0, |-h|=h, we have
f(1-h)=h
Also,
f(1)=0
Therefore,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{h-0}{-h}
\displaystyle =\lim_{h\to0}(-1)
i.e. LHD \displaystyle =-1 \qquad ...(2)
From (1) and (2),
\displaystyle \text{RHD}=1\ne -1=\text{LHD}
Since the Left-Hand Derivative and Right-Hand Derivative are not equal, the derivative does not exist at x=1.
Hence, f(x)=|x-1| is not differentiable at x=1.
Note: Modulus functions are not differentiable at points where the graph has a sharp corner or cusp. However, they are differentiable at all other points where the Left-Hand Derivative and Right-Hand Derivative are equal.
Question 10: G. I. Function is Non-Differentiable?
10. Prove that the greatest integer function defined by
f(x)=[x],\ 0\lt x\lt 3
is not differentiable at x=1 and x=2.
Solution
We shall show that the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) are not equal at x=1 and x=2.
At x=1
Right-Hand Derivative (RHD)
Using the definition,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}
Since \ 0\lt h\lt 1
So, f(1+h)=[1+h]=1
Also,
f(1)=[1]=1
Therefore,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{1-1}{h}
\displaystyle =0 \qquad ...(1)
Left-Hand Derivative (LHD)
Using the definition,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}
Since \ 0\lt h\lt 1
f(1-h)=[1-h]=0
Also,
f(1)=1
Therefore,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{0-1}{-h}
\displaystyle =\lim_{h\to0}\frac{1}{h}
Since \displaystyle \lim_{h\to0}\frac{1}{h}=\infty, LHD does not exist as a finite value.
Thus,
\displaystyle \text{LHD}\ne \text{RHD}
Therefore, f(x) is not differentiable at x=1.
At x=2
Right-Hand Derivative (RHD)
Using the definition,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}
Since \ 0\lt h\lt 1
f(2+h)=[2+h]=2
Also,
f(2)=[2]=2
Therefore,
\displaystyle \text{RHD}=\lim_{h\to0}\frac{2-2}{h}
\displaystyle =0 \qquad ...(2)
Left-Hand Derivative (LHD)
Using the definition,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(2-h)-f(2)}{-h}
Since \ 0\lt x\lt 3
f(2-h)=[2-h]=1
Also,
f(2)=2
Therefore,
\displaystyle \text{LHD}=\lim_{h\to0}\frac{1-2}{-h}
\displaystyle =\lim_{h\to0}\frac{1}{h}
Since \displaystyle \lim_{h\to0}\frac{1}{h}=\infty, LHD does not exist as a finite value.
Thus,
\displaystyle \text{LHD}\ne \text{RHD}
Therefore, f(x) is not differentiable at x=2.
Hence, the greatest integer function f(x)=[x] is not differentiable at x=1 and x=2.
Note: The greatest integer function is not differentiable at integer values because it has jump discontinuities there. However, on every open interval between two consecutive integers, the function is constant and hence differentiable.
Common Mistakes to Avoid
- Forgetting the Chain Rule: Students often differentiate only the outer function and forget to multiply by the derivative of the inner function. For example, \frac{d}{dx}(\sin x^2)\ne\cos(x^2). The correct derivative is 2x\cos(x^2).
- Differentiating composite functions incorrectly: In expressions like \cos(\sin x) or \sin(x^2+1), identify the outer and inner functions before applying the Chain Rule.
- Ignoring the derivative of the inner function: While differentiating (2x+1)^3, students often write 3(2x+1)^2 and forget to multiply by \frac{d}{dx}(2x+1)=2.
- Using Product Rule incorrectly: For uv, the derivative is not \frac{du}{dx}\cdot\frac{dv}{dx}. The correct formula is \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.
- Using Quotient Rule incorrectly: Students often confuse the order of terms in the numerator. Remember the shortcut \frac{DN'-ND'}{D^2}, where N and D denote the numerator and denominator respectively.
- Assuming continuity implies differentiability: Every differentiable function is continuous, but the converse is not true. A function may be continuous yet not differentiable at a point.
- Not checking both LHD and RHD: To prove differentiability at x=c, both the Left-Hand Derivative and Right-Hand Derivative must exist and be equal.
- Confusing LHD and RHD formulas: Remember that
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(c-h)-f(c)}{-h}
- Forgetting to simplify before taking limits: While finding LHD and RHD, simplify the expression first and then evaluate the limit. This often makes the calculation much easier.
- Assuming absolute value functions are differentiable everywhere: Functions such as f(x)=|x| are continuous everywhere but not differentiable at x=0 because LHD and RHD are unequal.
- Ignoring nested functions: In expressions like \sec(\tan\sqrt{x}), the Chain Rule must be applied repeatedly from the outermost function to the innermost function.
- Differentiating trigonometric functions incorrectly: Common sign errors include writing \frac{d}{dx}(\cos x)=\sin x or \frac{d}{dx}(\cot x)=\csc^2x. Remember the correct derivatives carefully.
- Not learning standard derivatives: Derivatives such as \frac{d}{dx}(x^n), \frac{d}{dx}(\sin x), \frac{d}{dx}(\cos x), \frac{d}{dx}(e^x) and \frac{d}{dx}(\ln x) should be memorised to avoid unnecessary mistakes.
Continue Learning
After completing Exercise 5.2 of Continuity and Differentiability, you should now be comfortable with the concept of differentiability, checking differentiability using Left-Hand and Right-Hand Derivatives, applying the basic rules of differentiation and differentiating composite functions using the Chain Rule.
To strengthen your understanding further, make sure that you revise:
- Meaning and definition of differentiability
- Derivative as a limit and its interpretation
- Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD)
- Conditions for differentiability at a point
- Relationship between continuity and differentiability
- Examples of functions that are continuous but not differentiable
- Sum, difference, product and quotient rules of differentiation
- Constant multiple rule and derivatives of standard functions
- Identifying composite functions and their inner and outer functions
- Applying the Chain Rule to composite and nested functions
- Differentiating trigonometric and algebraic composite functions
- Using multiple differentiation rules together in a single problem
Explore More
Try differentiating a variety of composite functions on your own and identify the inner and outer functions before applying the Chain Rule. Also practise questions involving Product Rule, Quotient Rule and differentiability using LHD and RHD. 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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