12th NCERT Continuity and Differentiability Exercise 5.2 Questions 10
Hi

Differentiate the function with respect to x

Question (1)

sin(x2+5)

Solution

y = sin(x2+5)
Differentiate w.r.t. x
\[\frac{d}{{dx}}\sin x = \cos x\] \[\frac{d}{{dx}}{x^n} = n{x^{n + 1}}\]
\[\frac{{dy}}{{dx}} = \cos \left( {{x^2} + 5} \right) \cdot \frac{d}{{dx}}\left( {{x^2} + 5} \right)\] \[\frac{{dy}}{{dx}} = \cos \left( {{x^2} + 5} \right) \cdot \left( {2x} \right)\] \[\frac{{dy}}{{dx}} = 2x\cos \left( {{x^2} + 5} \right)\]

Question (2)

cos(sinx)

Solution

y = cos(sinx)
Differentiate w.r.t. x
\[\frac{d}{{dx}}\cos x = - \sin x\] \[\frac{d}{{dx}}\sin x = \cos x\]
\[\frac{{dy}}{{dx}} = - \sin \left( {\sin x} \right) \cdot \frac{d}{{dx}}\sin x\] \[\frac{{dy}}{{dx}} = - \sin \left( {\sin x} \right) \cdot \cos x\] \[\frac{{dy}}{{dx}} = - \cos x\sin \left( {\sin x} \right)\]

Question (3)

sin(ax+b)

Solution

y = sin(ax+b)
Differentiate w.r.t. x
\[\frac{d}{{dx}}\sin x = \cos x\] \[\frac{d}{{dx}}ax = a\] \[\frac{d}{{dx}}\left( {constant} \right) = 0\]
\[\frac{{dy}}{{dx}} = \cos \left( {ax + b} \right)\frac{d}{{dx}}\left( {ax + b} \right)\] \[\frac{{dy}}{{dx}} = \cos \left( {ax + b} \right)a\] \[\frac{{dy}}{{dx}} = a\cos \left( {ax + b} \right)\]

Question (4)

sec(tan√x)

Solution

y = sec(tan√x) Differentiate w.r.t. x
\[\frac{d}{{dx}}\sec x = \sec x\tan x\] \[\frac{d}{{dx}}\tan x = {\sec ^2}x\] \[\frac{d}{{dx}}\sqrt x = \frac{1}{{2\sqrt x }}\]
\[\frac{{dy}}{{dx}} = \sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right)\frac{d}{{dx}}\left( {\tan \sqrt x } \right)\] \[\frac{{dy}}{{dx}} = \sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right){\sec ^2}\sqrt x \frac{d}{{dx}}\sqrt x \] \[\frac{{dy}}{{dx}} = \sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right){\sec ^2}\sqrt x \frac{1}{{2\sqrt x }}\] \[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\sec \left( {\tan \sqrt x } \right)\tan \left( {\tan \sqrt x } \right){\sec ^2}\sqrt x \]

Question (5)

\[\frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}\]

Solution

\[y = \frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}\] Differentiate w.r.t. x
\[\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}\] \[\frac{d}{{dx}}\sin x = \cos x\] \[\frac{d}{{dx}}\cos x = - \sin x\]
\[\frac{{dy}}{{dx}} = \frac{{\cos \left( {cx + d} \right)\frac{d}{{dx}}\sin \left( {ax + b} \right) - \sin \left( {ax + b} \right)\frac{d}{{dx}}\cos \left( {cx + d} \right)}}{{{{\left[ {\cos \left( {cx + d} \right)} \right]}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)a - \sin \left( {ax + b} \right)\left[ { - \sin \left( {cx + d} \right)} \right]c}}{{{{\left[ {\cos \left( {cx + d} \right)} \right]}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{a\cos \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}} + \frac{{c\sin \left( {ax + b} \right)\sin \left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}\] \[\frac{{dy}}{{dx}} = a\cos \left( {ax + b} \right)\sec \left( {cx + d} \right) + c\sin \left( {ax + b} \right)\tan \left( {cx + d} \right)\sec \left( {cx + d} \right)\]

Question (6)

cosx3 sin2(x5)

Solution

y = cosx3 sin2(x5) Differentiate w.r.t. x
\[\frac{d}{{dx}}uv = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}\] \[\frac{d}{{dx}}{\sin ^2}x = 2\sin x\cos x\] \[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
\[\frac{{dy}}{{dx}} = \cos {x^3}\frac{d}{{dx}}{\sin ^2}\left( {{x^5}} \right) + {\sin ^2}{x^5}\frac{d}{{dx}}\cos \left( {{x^3}} \right)\] \[\frac{{dy}}{{dx}} = \cos {x^3}2\sin \left( {{x^5}} \right)\cos {x^5}\frac{d}{{dx}}\left( {{x^5}} \right) + {\sin ^2}{x^5}\left[ { - \sin {x^3}} \right]\frac{d}{{dx}}\left( {{x^3}} \right)\] \[\frac{{dy}}{{dx}} = 2\sin \left( {{x^5}} \right)\cos {x^5}\cos {x^3}\left( {5{x^4}} \right) - {\sin ^2}{x^5}\sin {x^3}\left( {3{x^2}} \right)\] \[\frac{{dy}}{{dx}} = 10{x^4}\sin {x^5}\cos {x^5}\cos {x^3} - 3{x^2}{\sin ^2}{x^5}\sin {x^3}\]

Question (7)

\[2\sqrt {\cot {x^2}} \]

Solution

\[y = 2\sqrt {\cot {x^2}} \] Differentiate w.r.t. x
\[\frac{d}{{dx}}\sqrt x = \frac{1}{{2\sqrt x }}\] \[\frac{d}{{dx}}\cot x = - cosec^2x\] \[\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
\[\frac{{dy}}{{dx}} =\require{cancel} \cancel{2}\frac{1}{{\cancel{2}\sqrt {\cot {x^2}} }}\frac{d}{{dx}}\cot {x^2}\] \[\frac{{dy}}{{dx}} = \frac{{ - {{cosec }^2}{x^2}}}{{\sqrt {\cot {x^2}} }} \times \frac{d}{{dx}}{x^2}\] \[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\sin }^2}{x^2}}} \times \frac{1}{{\sqrt {\frac{{\cos {x^2}}}{{\sin {x^2}}}} }}2x\] \[\frac{{dy}}{{dx}} = \frac{{ - 2x}}{{{{\sin }^{\frac{3}{2}}}{x^2}\sqrt {\cos {x^2}} }}\] \[\frac{{dy}}{{dx}} = \frac{{ - 2x}}{{\sin {x^2}\sqrt {\sin {x^2}} \sqrt {\cos {x^2}} }}\] Multiply by √2 \[\frac{{dy}}{{dx}} = \frac{{ - 2x\sqrt 2 }}{{\sin {x^2}\sqrt {2\sin {x^2}\cos {x^2}} }}\] \[\frac{{dy}}{{dx}} = \frac{{ - 2\sqrt 2 x}}{{\sin {x^2}\sqrt {\sin 2{x^2}} }}\]

Question (8)

\[\cos \left( {\sqrt x } \right)\]

Solution

\[y = \cos \left( {\sqrt x } \right)\] Differentiate w.r.t. x
\[\frac{d}{{dx}}\cos x = - \sin x\] \[\frac{d}{{dx}}\sqrt x = \frac{1}{{2\sqrt x }}\]
\[\frac{{dy}}{{dx}} = - \sin \left( {\sqrt x } \right)\frac{d}{{dx}}\sqrt x \] \[\frac{{dy}}{{dx}} = - \sin \left( {\sqrt x } \right)\frac{1}{{2\sqrt x }}\] \[\frac{{dy}}{{dx}} = \frac{{ - \sin \left( {\sqrt x } \right)}}{{2\sqrt x }}\]

Question (9)

Prove that the function "f" given by f(x) = |x-1|, x ∈ R is not differentiable at x=1

Solution

\[\text{If}\quad \mathop {\lim }\limits_{x \to {a^ - }} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} = \mathop {\lim }\limits_{x \to {a^ + }} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\] then f is differentiable at x = a
|x-1|= x-1, x ≥ 1
|x-1|= -(x-1), x < 1
f(x) = |x-1|, x ∈ R f(x) = x-1, x ≥ 1
f(x) = -(x-1), x < 1
Now, f(x) = |1-1| = 0
\[\text{LHS=} \quad \mathop {\lim }\limits_{x \to {1^ - }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{ - \left( {x - 1} \right) - 0}}{{x - 1}} = - 1\] \[\text{RHS=} \quad \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1 - 0}}{{x - 1}} = 1\] \[\because \mathop {\lim }\limits_{x \to {1^ - }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} \ne \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\] f is not differentiable at x=1

Question (10)

Prove that the greatest integer function defined by f(x) =[x] 0 < x < 3 is not differentiable at x=1 and x=2

Solution

\[\text{If}\quad \mathop {\lim }\limits_{x \to {a^ - }} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} - \mathop {\lim }\limits_{x \to {a^ + }} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\] then f is differentiable at x = a
as x ⇒ a- [x] = a-1
as x ⇒ a+ [x] = a
Let us check for x=1
∴ f(1) = [1] = 1 \[\text{LHL} \quad\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\left[ x \right] - 1}}{{x - 1}} = \frac{{0 - 1}}{{1 - 1}} = -\infty\] as x ⇒ 1 x -1 ≠ 0 \[\text{RHL} \quad\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\left[ x \right] - 1}}{{x - 1}} = \frac{{1 - 1}}{{1 - 1}} = 0\] as x ⇒ 1 x -1 ≠ 0 f is not differentiable at x= 1 Let us check for x=2 ∴ f(2) = [2] = 2 \[\text{LHL} \quad\mathop {\lim }\limits_{x \to {2^ - }} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left[ x \right] - 2}}{{x - 2}} = \frac{{-1 - 2}}{{2 - 2}} = - \infty \] \[\text{RHL} \quad\mathop {\lim }\limits_{x \to {2^ + }} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left[ x \right] - 2}}{{x - 2}} = \frac{{2 - 2}}{{2 - 2}} = 0\] f is not differentiable at x= 2
Exercise 5.1⇐
⇒Exercise 5.3