NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.2
Note: The Chain Rule is used to differentiate composite functions. If \( y = f(g(x)) \), let \( u = g(x) \), then \( y = f(u) \). The derivative is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Q1
Differentiate the function with respect to \( x \): \( \sin(x^2 + 5) \)
Solution:
Let \( y = \sin(x^2 + 5) \).
Using the chain rule:
\[ \begin{aligned} \frac{dy}{dx} &= \cos(x^2 + 5) \cdot \frac{d}{dx}(x^2 + 5) \\ &= \cos(x^2 + 5) \cdot (2x) \\ &= 2x \cos(x^2 + 5) \end{aligned} \]
Q2
Differentiate the function with respect to \( x \): \( \cos(\sin x) \)
Solution:
Let \( y = \cos(\sin x) \).
Using the chain rule:
\[ \begin{aligned} \frac{dy}{dx} &= -\sin(\sin x) \cdot \frac{d}{dx}(\sin x) \\ &= -\sin(\sin x) \cdot \cos x \\ &= -\cos x \sin(\sin x) \end{aligned} \]
Q3
Differentiate the function with respect to \( x \): \( \sin(ax + b) \)
Solution:
Let \( y = \sin(ax + b) \).
\[ \begin{aligned} \frac{dy}{dx} &= \cos(ax + b) \cdot \frac{d}{dx}(ax + b) \\ &= \cos(ax + b) \cdot a \\ &= a \cos(ax + b) \end{aligned} \]
Q4
Differentiate the function with respect to \( x \): \( \sec(\tan(\sqrt{x})) \)
Solution:
Let \( y = \sec(\tan(\sqrt{x})) \).
Apply chain rule repeatedly:
\[ \begin{aligned} \frac{dy}{dx} &= \sec(\tan\sqrt{x})\tan(\tan\sqrt{x}) \cdot \frac{d}{dx}(\tan\sqrt{x}) \\ &= \sec(\tan\sqrt{x})\tan(\tan\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \\ &= \sec(\tan\sqrt{x})\tan(\tan\sqrt{x}) \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sec(\tan\sqrt{x})\tan(\tan\sqrt{x})\sec^2(\sqrt{x})}{2\sqrt{x}} \end{aligned} \]
Q5
Differentiate the function with respect to \( x \): \( \frac{\sin(ax + b)}{\cos(cx + d)} \)
Solution:
Use the Quotient Rule: \( \left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2} \).
Let \( u = \sin(ax+b) \) and \( v = \cos(cx+d) \).
\( u' = a\cos(ax+b) \).
\( v' = -c\sin(cx+d) \).
\[ \begin{aligned} \frac{dy}{dx} &= \frac{\cos(cx+d) \cdot a\cos(ax+b) - \sin(ax+b) \cdot (-c\sin(cx+d))}{\cos^2(cx+d)} \\ &= \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)} \\ &= a\cos(ax+b)\sec(cx+d) + c\sin(ax+b)\tan(cx+d)\sec(cx+d) \end{aligned} \]
Q6
Differentiate the function with respect to \( x \): \( \cos x^3 \cdot \sin^2(x^5) \)
Solution:
Use the Product Rule: \( (uv)' = u'v + uv' \).
Let \( u = \cos x^3 \) and \( v = \sin^2(x^5) \).
\[ \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}(\cos x^3) \cdot \sin^2(x^5) + \cos x^3 \cdot \frac{d}{dx}(\sin^2(x^5)) \\ &= (-\sin x^3 \cdot 3x^2) \sin^2(x^5) + \cos x^3 (2\sin(x^5) \cdot \cos(x^5) \cdot 5x^4) \\ &= -3x^2 \sin x^3 \sin^2(x^5) + 10x^4 \sin(x^5) \cos(x^5) \cos x^3 \end{aligned} \]
Q7
Differentiate the function with respect to \( x \): \( 2\sqrt{\cot(x^2)} \)
Solution:
Let \( y = 2\sqrt{\cot(x^2)} \).
\[ \begin{aligned} \frac{dy}{dx} &= 2 \cdot \frac{1}{2\sqrt{\cot(x^2)}} \cdot \frac{d}{dx}(\cot(x^2)) \\ &= \frac{1}{\sqrt{\cot(x^2)}} \cdot (-\text{cosec}^2(x^2)) \cdot \frac{d}{dx}(x^2) \\ &= \frac{-\text{cosec}^2(x^2) \cdot 2x}{\sqrt{\cot(x^2)}} \\ &= \frac{-2x}{\sin^2(x^2)\sqrt{\frac{\cos(x^2)}{\sin(x^2)}}} = \frac{-2x}{\sin(x^2)\sqrt{\sin(x^2)\cos(x^2)}} \\ &= \frac{-2\sqrt{2}x}{\sin(x^2)\sqrt{2\sin(x^2)\cos(x^2)}} = \frac{-2\sqrt{2}x}{\sin(x^2)\sqrt{\sin(2x^2)}} \end{aligned} \]
Q8
Differentiate the function with respect to \( x \): \( \cos(\sqrt{x}) \)
Solution:
Let \( y = \cos(\sqrt{x}) \).
\[ \begin{aligned} \frac{dy}{dx} &= -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \\ &= -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \\ &= \frac{-\sin(\sqrt{x})}{2\sqrt{x}} \end{aligned} \]
Q9
Prove that the function \( f \) given by \( f(x) = |x - 1|, x \in \mathbf{R} \) is not differentiable at \( x = 1 \).
Solution:
A function is differentiable at \( x = c \) if LHD = RHD at \( c \).
Let's find the Left Hand Derivative (LHD) at \( x = 1 \):
\[ \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{|(1+h)-1| - 0}{h} = \lim_{h \to 0} \frac{|h|}{h} \]
Since \( h < 0 \), \( |h| = -h \). So LHD = \( \frac{-h}{h} = -1 \).
Right Hand Derivative (RHD) at \( x = 1 \):
\[ \lim_{h \to 0^+} \frac{|h|}{h} = \frac{h}{h} = 1 \]
Since LHD \( \neq \) RHD, \( f(x) \) is not differentiable at \( x = 1 \).
Q10
Prove that the greatest integer function defined by \( f(x) = [x], 0 < x < 3 \) is not differentiable at \( x = 1 \) and \( x = 2 \).
Solution:
At \( x = 1 \):
- LHD: \( \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} = \frac{[1-h] - 1}{-h} = \frac{0 - 1}{-h} = \frac{1}{h} \to \infty \).
- RHD: \( \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \frac{[1+h] - 1}{h} = \frac{1 - 1}{h} = 0 \).
Since derivatives do not exist (infinite) or are unequal, it is not differentiable at \( x = 1 \).
At \( x = 2 \):
- LHD: \( \lim_{h \to 0} \frac{f(2-h) - f(2)}{-h} = \frac{1 - 2}{-h} = \frac{-1}{-h} \to \infty \).
- RHD: \( \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \frac{2 - 2}{h} = 0 \).
Not differentiable at \( x = 2 \).