NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Use the quotient rule: if \(\displaystyle y=\dfrac{u}{v}\), then \(\displaystyle \dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\), valid wherever \(\displaystyle v\neq 0\).Here \(\displaystyle u=e^{x}\) and \(\displaystyle v=\sin x\), so \(\displaystyle u'=e^{x}\) and \(\displaystyle v'=\cos x\). Hence \[\frac{d}{dx}\left(\frac{e^{x}}{\sin x}\right)=\frac{e^{x}\sin x-e^{x}\cos x}{\sin^{2}x}=\frac{e^{x}\left(\sin x-\cos x\right)}{\sin^{2}x}.\]The function itself is undefined where \(\displaystyle \sin x=0\), so the derivative is \[\frac{e^{x}(\sin x-\cos x)}{\sin^{2}x},\qquad x\neq n\pi,\ n\in\mathbf{Z}.\]

Chain rule with the outer function \(\displaystyle e^{t}\) and inner \(\displaystyle t=\sin^{-1}x\): \[\frac{d}{dx}\left(e^{t}\right)=e^{t}\cdot\frac{dt}{dx}.\]The standard derivative of the inverse sine is \(\displaystyle \dfrac{d}{dx}\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^{2}}}\), which itself holds only for \(\displaystyle -1<x<1\) (at \(\displaystyle x=\pm 1\) the tangent is vertical). Therefore \[\frac{d}{dx}\left(e^{\sin^{-1}x}\right)=e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^{2}}}.\]So the derivative is \(\displaystyle \dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\), for \(\displaystyle x\in(-1,1)\).

Chain rule, outer function \(\displaystyle e^{t}\) with \(\displaystyle t=x^{3}\): \[\frac{d}{dx}\left(e^{x^{3}}\right)=e^{x^{3}}\cdot\frac{d}{dx}\left(x^{3}\right)=e^{x^{3}}\cdot 3x^{2}.\]Note the exponent is \(\displaystyle x^{3}\), not \(\displaystyle 3x\); the factor brought down is \(\displaystyle 3x^{2}\), not \(\displaystyle 3\).Hence \(\displaystyle \dfrac{d}{dx}\left(e^{x^{3}}\right)=3x^{2}e^{x^{3}}\), for all \(\displaystyle x\in\mathbf{R}\).

This is a chain of three functions: \(\displaystyle \sin\) of \(\displaystyle \tan^{-1}\) of \(\displaystyle e^{-x}\). Differentiate from the outside in, using \(\displaystyle \dfrac{d}{dx}\left(\tan^{-1}t\right)=\dfrac{1}{1+t^{2}}\).With \(\displaystyle y=\sin\left(\tan^{-1}e^{-x}\right)\), \[\frac{dy}{dx}=\cos\left(\tan^{-1}e^{-x}\right)\cdot\frac{1}{1+\left(e^{-x}\right)^{2}}\cdot\frac{d}{dx}\left(e^{-x}\right).\]The innermost derivative carries the minus sign: \(\displaystyle \dfrac{d}{dx}\left(e^{-x}\right)=-e^{-x}\). So \[\frac{dy}{dx}=\frac{-e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}.\]One may simplify the cosine: if \(\displaystyle \theta=\tan^{-1}e^{-x}\) then \(\displaystyle \tan\theta=e^{-x}\) with \(\displaystyle \theta\in\left(0,\tfrac{\pi}{2}\right)\), so \(\displaystyle \cos\theta=\dfrac{1}{\sqrt{1+e^{-2x}}}\), giving the equivalent form \[\frac{dy}{dx}=\frac{-e^{-x}}{\left(1+e^{-2x}\right)^{3/2}},\qquad x\in\mathbf{R}.\]

Chain rule twice, with \(\displaystyle \dfrac{d}{dx}\left(\log t\right)=\dfrac{1}{t}\) (natural logarithm).Let \(\displaystyle y=\log\left(\cos e^{x}\right)\). Then \[\frac{dy}{dx}=\frac{1}{\cos e^{x}}\cdot\frac{d}{dx}\left(\cos e^{x}\right)=\frac{1}{\cos e^{x}}\cdot\left(-\sin e^{x}\right)\cdot\frac{d}{dx}\left(e^{x}\right).\]The last factor is \(\displaystyle e^{x}\) itself, so \[\frac{dy}{dx}=-\frac{e^{x}\sin e^{x}}{\cos e^{x}}=-e^{x}\tan\left(e^{x}\right).\]Thus the derivative is \(\displaystyle -e^{x}\tan\left(e^{x}\right)\), valid on the set where the function is defined, i.e. where \(\displaystyle \cos e^{x}>0\).

Differentiate term by term (the derivative of a finite sum is the sum of the derivatives), each term by the chain rule \(\displaystyle \dfrac{d}{dx}\left(e^{x^{n}}\right)=e^{x^{n}}\cdot nx^{n-1}\).Let \(\displaystyle y=e^{x}+e^{x^{2}}+e^{x^{3}}+e^{x^{4}}+e^{x^{5}}\). Then \[\frac{dy}{dx}=e^{x}\cdot 1+e^{x^{2}}\cdot 2x+e^{x^{3}}\cdot 3x^{2}+e^{x^{4}}\cdot 4x^{3}+e^{x^{5}}\cdot 5x^{4}.\]So \[\frac{dy}{dx}=e^{x}+2x\,e^{x^{2}}+3x^{2}e^{x^{3}}+4x^{3}e^{x^{4}}+5x^{4}e^{x^{5}},\qquad x\in\mathbf{R}.\]

First rewrite the surd as a single exponential, using \(\displaystyle \sqrt{a}=a^{1/2}\) and \(\displaystyle \left(e^{t}\right)^{1/2}=e^{t/2}\): \[y=\sqrt{e^{\sqrt{x}}}=e^{\frac{\sqrt{x}}{2}}=e^{\frac{1}{2}x^{1/2}}.\]Now the chain rule with \(\displaystyle t=\tfrac{1}{2}x^{1/2}\), for which \(\displaystyle \dfrac{dt}{dx}=\tfrac{1}{2}\cdot\tfrac{1}{2}x^{-1/2}=\dfrac{1}{4\sqrt{x}}\) (this is the step most often dropped — the \(\displaystyle \tfrac12\) from the outer square root and the \(\displaystyle \tfrac12\) from \(\displaystyle \sqrt{x}\) both appear): \[\frac{dy}{dx}=e^{\frac{\sqrt{x}}{2}}\cdot\frac{1}{4\sqrt{x}}.\]Hence \[\frac{dy}{dx}=\frac{e^{\sqrt{x}/2}}{4\sqrt{x}}=\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}},\qquad x>0.\]

Chain rule with the outer function \(\displaystyle \log t\), \(\displaystyle t=\log x\), using \(\displaystyle \dfrac{d}{dx}\left(\log x\right)=\dfrac{1}{x}\).Let \(\displaystyle y=\log\left(\log x\right)\). Then \[\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{d}{dx}\left(\log x\right)=\frac{1}{\log x}\cdot\frac{1}{x}.\]The restriction \(\displaystyle x>1\) is what makes \(\displaystyle \log x>0\), so that the outer logarithm is defined and \(\displaystyle \log x\neq 0\) in the denominator.Therefore \[\frac{dy}{dx}=\frac{1}{x\log x},\qquad x>1.\]

Quotient rule with \(\displaystyle u=\cos x\), \(\displaystyle v=\log x\), so \(\displaystyle u'=-\sin x\) and \(\displaystyle v'=\dfrac{1}{x}\): \[\frac{d}{dx}\left(\frac{\cos x}{\log x}\right)=\frac{u'v-uv'}{v^{2}}=\frac{\left(-\sin x\right)\log x-\cos x\cdot\dfrac{1}{x}}{\left(\log x\right)^{2}}.\]Multiply numerator and denominator by \(\displaystyle x\) to clear the internal fraction: \[=\frac{-x\sin x\,\log x-\cos x}{x\left(\log x\right)^{2}}=-\frac{x\log x\,\sin x+\cos x}{x\left(\log x\right)^{2}}.\]So the derivative is \[-\frac{x\log x\,\sin x+\cos x}{x\left(\log x\right)^{2}},\qquad x>0,\ x\neq 1\ \text{(since }\log 1=0\text{)}.\]

Chain rule with the outer function \(\displaystyle \cos t\), \(\displaystyle t=\log x+e^{x}\), so that \(\displaystyle \dfrac{dy}{dx}=-\sin t\cdot\dfrac{dt}{dx}\).Differentiate the inner sum term by term: \[\frac{dt}{dx}=\frac{d}{dx}\left(\log x\right)+\frac{d}{dx}\left(e^{x}\right)=\frac{1}{x}+e^{x}.\]Hence \[\frac{d}{dx}\Big(\cos\left(\log x+e^{x}\right)\Big)=-\sin\left(\log x+e^{x}\right)\left(\frac{1}{x}+e^{x}\right),\qquad x>0,\] the restriction \(\displaystyle x>0\) being needed for \(\displaystyle \log x\) to exist.