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Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.4
  • Last Updated : 18 Mar, 2021

Differentiate the following w.r.t x

Question 1. y = \frac{e^x}{\sin x}

Solution:

\frac{dy}{dx}=\frac{d}{dx}(\frac{e^x}{\sin x})

\frac{dy}{dx}=\frac{\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin x}{sin^2x}          (\frac{d}{dx}(\frac{u}{v})=v\frac{\frac{du}{dx}-u\frac{dv}{dx}}{v^2})

\frac{dy}{dx}=\frac{\sin x e^x-e^x\cos x}{sin^2x}

\frac{dy}{dx}=e^x\frac{(\sin x-\cos x)}{\sin^2x}



Question 2. y=e^{\sin^{-1}x}

Solution:

\frac{dy}{dx}=\frac{d}{dx}(e^{sin^{-1}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}}

Question 3. y=e^{x^{3}}

Solution:

\frac{dy}{dx}=\frac{d}{dx}(e^{x^{3}})

\frac{dy}{dx}=e^{x^{3}}.\frac{dy}{dx}(x^3)=\frac{dy}{dx}=e^{x^{3}}.(3x^{2})

\frac{dy}{dx}=3x^2e^{x^3}

Question 4. y=sin(tan-1e-x)

Solution:



\frac{dy}{dx}=\frac{d}{dx}(\sin(\tan^{-1}e^{-x}))

\frac{dy}{dx}=\cos(\tan^{-1}e^{-x}).\frac{d}{dx}(\tan^{-1}e^{-x})

\frac{dy}{dx}=\cos(\tan^{-1}e^{-x}).\frac{1}{1+e^{-2x}}.\frac{d}{dx}e^{-x}

\frac{dy}{dx}=\frac{-\cos(tan^{-1}e^{-x})}{1+e^{-2x}}.e^{-x}

Question 5. y = log(cos ex)

Solution:

\frac{dy}{dx}=\frac{d}{dx}(\log(\cos e^x))

\frac{dy}{dx}=\frac{1}{\cos e^x}.\frac{d}{dx}.\cos e^x

\frac{dy}{dx}=\frac{1}{\cos e^x}.(-\sin e^x).\frac{d}{dx}e^x

\frac{dy}{dx}=\frac{-\sin e^x}{\cos e^x}.e^x

\frac{dy}{dx}=-e^x\tan e^x

Question 6. y = e^{x} + e^{x^{2}} + e^{x^{3}} + e^{x^{4}} + e^{x^{5}}

Solution:

\frac{dy}{dx}=\frac{d}{dx}e^x+\frac{d}{dx}e^{x^{2}}+\frac{d}{dx}e^{x^{3}}+\frac{d}{dx}e^{x^{4}}+\frac{d}{dx}e^{x^{5}}

\frac{dy}{dx}=e^x+e^{x^{2}}.\frac{d}{dx}x^2+e^{x^{3}}\frac{d}{dx}x^3+e^{x^{4}}\frac{d}{dx}x^4+e^{x^{5}}\frac{d}{dx}x^5

\frac{dy}{dx}=e^x2x^{x^{2}}+3x^2e^{x^{3}}+4x^3e^{x^{4}}+5x^4e^{x^{5}}

Question 7. y=\sqrt{e^{\sqrt{x}}}

Solution:

\frac{dy}{dx}=\frac{d}{dx}\sqrt{e^{\sqrt{x}}}

\frac{dy}{dx}=\frac{1}{2}(e^{\sqrt{x}})^{-1/2}.\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{\sqrt{e^{\sqrt{x}}}}{2}.\frac{1}{2\sqrt{x}}

\frac{dy}{dx}=\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}

Question 8. y = log(log x)

Solution:

\frac{dy}{dx}=\frac{d}{dx}(\log (\log x))

\frac{dy}{dx}=\frac{1}{\log x}.\frac{d}{dx}\log x

\frac{dy}{dx}=\frac{1}{x\log x}

Question 9. y = \frac{\cos x}{\log x}

Solution:

\frac{dy}{dx}=\frac{\log x\frac{d}{dx}\cos x-\cos x\frac{dy}{dx}\log x}{\log^2x}                                          (\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2})

\frac{dy}{dx}=\frac{\log x(-\sin x)-\cos x.(\frac{1}{x})}{\log^2x}

\frac{dy}{dx}=\frac{-\sin x}{\log x}-\frac{\cos x}{x\log^2x}

Question 10. y= cos (log x+ex)

Solution:

\frac{dy}{dx}=\frac{d}{dx}(\cos(\log x+e^x))

\frac{dy}{dx}=-\sin(\log x+e^x).\frac{d}{dx}\sin(\log x+e^x)

\frac{dy}{dx}=-\sin(\log x+e^x).(\frac{1}{x}+e^x)

\frac{dy}{dx}=\frac{-\sin(\log x+e^x)(1+xe^x)}{x}

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