12th NCERT Continuity and Differentiability Exercise 5.4 Questions 10

Differentiate the following w.r.t x

Question (1)

\[\frac{{{e^x}}}{{\sin x}}\]

Solution

\[\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}\] \[\frac{d}{{dx}}{e^x} = {e^x}\] \[\frac{d}{{dx}}\sin x = \cos x\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \frac{{\sin x\frac{d}{{dx}}{e^x} - {e^x}\frac{d}{{dx}}\sin x}}{{{{\left( {\sin x} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\sin x{e^x} - {e^x}\cos x}}{{{{\sin }^2}x}}\] \[\frac{{dy}}{{dx}} = \frac{{{e^x}\left( {\sin x - \cos x} \right)}}{{{{\sin }^2}x}}\]

Question (2)

\[{e^{{{\sin }^{ - 1}}x}}\]

Solution

\[\frac{d}{{dx}}{e^x} = {e^x}\] \[\frac{d}{{dx}}{\sin ^{ - 1}}x = \frac{1}{{\sqrt {1 - {x^2}} }}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}}\frac{d}{{dx}}{\sin ^{ - 1}}x\] \[\frac{{dy}}{{dx}} = {e^{{{\sin }^{ - 1}}x}}\frac{1}{{\sqrt {1 - {x^2}} }}\] \[\frac{{dy}}{{dx}} = \frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}\]

Question (3)

\[{e^{{x^3}}}\]

Solution

\[\frac{d}{{dx}}{e^x} = {e^x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = {e^{{x^3}}} \cdot \frac{d}{{dx}}{x^3}\] \[\frac{{dy}}{{dx}} = {e^{{x^3}}} \cdot 3{x^2}\] \[\frac{{dy}}{{dx}} = 3{x^2}{e^{{x^3}}}\]

Question (4)

sin (tan- e-x)

Solution

Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{d}{{dx}}{\tan ^{ - 1}}\left( {{e^{ - x}}} \right)\] \[\frac{{dy}}{{dx}} = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}{e^{ - x}}\] \[\frac{{dy}}{{dx}} = \frac{{\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{-2x}}}}{e^{ - x}} \times \left( { - 1} \right)\] \[\frac{{dy}}{{dx}} = \frac{{ - {e^x}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{-2x}}}}\]

Question (5)

log(cosex)

Solution

\[\frac{d}{{dx}}\log x = \frac{1}{x}\] \[\frac{d}{{dx}}{e^x} = {e^x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}}\frac{d}{{dx}}\cos {e^x}\] \[\frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}} \times - \sin {e^x}\frac{d}{{dx}}{e^x}\] \[\frac{{dy}}{{dx}} = - \tan {e^x} \cdot {e^x}\] \[\frac{{dy}}{{dx}} = - {e^x}\tan {e^x}\]

Question (6)

ex + ex2+ ....+ ex5

Solution

Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = {e^x} + {e^{{x^2}}}\frac{d}{{dx}}{x^2} + ... + {e^{{x^5}}}\frac{d}{{dx}}{x^5}\] \[ = {e^x} + {e^{{x^2}}} \cdot 2x + {e^{{x^3}}} \cdot 3x + {e^{{x^4}}} \cdot 4x + {e^{{x^5}}} \cdot 5x\] \[ = {e^x} + 2x{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}{e^{{x^4}}} + 5{x^4}{e^{{x^5}}}\]

Question (7)

\[\sqrt {{e^{\sqrt x }}} ,x > 0\]

Solution

\[\frac{d}{{dx}}\sqrt x = \frac{1}{{2\sqrt x }}\] \[\frac{d}{{dx}}{e^x} = {e^x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{e^{\sqrt x }}} }}\frac{d}{{dx}}{e^{\sqrt x }}\] \[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{e^{\sqrt x }}} }}{e^{\sqrt x }}\frac{d}{{dx}}\sqrt x \] \[\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{2\sqrt {{e^{\sqrt x }}} }}\frac{1}{{2\sqrt x }}\] \[\frac{{dy}}{{dx}} = \frac{{\sqrt {{e^{\sqrt x }}} }}{{4\sqrt x }}\]

Question (8)

log (logx), x > 1

Solution

\[\frac{d}{{dx}}\log x = \frac{1}{x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \frac{1}{{\log x}}\frac{d}{{dx}}\left( {\log x} \right)\] \[\frac{{dy}}{{dx}} = \frac{1}{{\log x}} \cdot \frac{1}{x}\] \[\frac{{dy}}{{dx}} = \frac{1}{{x\log x}}\]

Question (9)

\[\frac{{\cos x}}{{\log x}},x > 0\]

Solution

\[\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}\] \[\frac{d}{{dx}}\cos x = - \sin x\] \[\frac{d}{{dx}}\log x = \frac{1}{x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = \frac{{\log x\frac{d}{{dx}}\cos x - \cos x\frac{d}{{dx}}\log x}}{{{{\left( {\log x} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\log x\left( { - \sin x} \right) - \cos x \cdot \frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - x\log x\sin x - \cos x}}{{x{{\left( {\log x} \right)}^2}}}\]

Question (10)

cos(logx+ex), x>0

Solution

\[\frac{d}{{dx}}\log x = \frac{1}{x}\]
Differentiating w.r.t. x \[\frac{{dy}}{{dx}} = - \sin \left( {\log x + {e^x}} \right)\frac{d}{{dx}}\left( {\log x + {e^x}} \right)\] \[\frac{{dy}}{{dx}} = - \sin \left( {\log x + {e^x}} \right)\frac{d}{{dx}}\left( {\log x + {e^x}} \right)\] \[\frac{{dy}}{{dx}} = - \left( {\frac{1}{x} + {e^x}} \right)\sin \left( {\log x + {e^x}} \right)\]
Exercise 5.3 ⇐
⇒ Exercise 5.5