# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

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## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.4) Exercise 5.4

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### NCERT Solutions for Class 12 Maths Chapter 5 Other Exercises

 Chapter 5 – Continuity and Differentiability Exercises Exercise 5.1 10 Short Questions and 24 Long Questions Exercise 5.2 2 Short Questions 8 Long Questions Exercise 5.3 9 short Questions and 6 long Questions Exercise 5.5 4 Short Questions and 14 Long Questions Exercise 5.6 1 Short Question and 1 Long Question Exercise 5.7 10 Short Questions and 7 Long Questions Exercise 5.8 Questions with Solutions

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Q.1

$\mathbf{\text{Differentiate}}\text{}\mathbf{\text{the}}\text{}\mathbf{\text{following}}\text{}\mathbf{\text{w}}\text{.}\mathbf{\text{r}}\text{.}\mathbf{\text{t}}\text{.}\mathbf{\text{x}}\text{:}\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{sinx}}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{sinx}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\frac{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{sinx}}\\ =\frac{\mathrm{sinx}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}}{{\left(\mathrm{sinx}\right)}^{2}}\left[\mathrm{By}\mathrm{Quotient}\mathrm{Rule}\right]\\ =\frac{\mathrm{sinx}.{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}}{{\left(\mathrm{sinx}\right)}^{2}}\\ =\frac{{\mathrm{e}}^{\mathrm{x}}\left(\mathrm{sinx}-\mathrm{cosx}\right)}{{\mathrm{sin}}^{2}\mathrm{x}}, \mathrm{x}\ne \mathrm{n\pi }, \mathrm{n}\in \mathbf{Z}\end{array}$

Q.2

$\mathbf{D}\mathbf{i}\mathbf{f}\mathbf{f}\mathbf{e}\mathbf{r}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{i}\mathbf{a}\mathbf{t}\mathbf{e}\text{}\mathbf{t}\mathbf{h}\mathbf{e}\text{}\mathbf{f}\mathbf{o}\mathbf{l}\mathbf{l}\mathbf{o}\mathbf{w}\mathbf{i}\mathbf{n}\mathbf{g}\text{}\mathbf{w}.\mathbf{r}.\mathbf{t}.\text{}\mathbf{x}:{e}^{si{n}^{-1}x}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}={\mathrm{e}}^{{\mathrm{sin}}^{-1}\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{sin}}^{-1}\mathrm{x}}\\ ={\mathrm{e}}^{{\mathrm{sin}}^{-1}\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{sin}}^{-1}\mathrm{x}\left[\mathrm{By}\mathrm{Chain}\mathrm{Rule}\right]\\ ={\mathrm{e}}^{{\mathrm{sin}}^{-1}\mathrm{x}}×\frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\\ =\frac{{\mathrm{e}}^{{\mathrm{sin}}^{-1}\mathrm{x}}}{\sqrt{1-{\mathrm{x}}^{2}}}, \mathrm{x}\in \left(-1,1\right)\\ \end{array}$

Q.3

$Differentiate\text{}the\text{}following\text{}w.r.t.\text{}x:\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{e}^{{x}^{3}}$

Ans

$Let y= e x 3 Differentiating both sides w.r.t. x, we get dy dx = d dx e x 3 = e x 3 d dx x 3 By Chain Rule = e x 3 ×3 x 2 =3 x 2 .e x 3 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabYeacaqGLbGaaeiDaiaabccacaqG5bGaeyypa0Jaaeyz amaaCaaaleqabaacbeGaa8hEamaaCaaameqabaGaa83maaaaaaaake aacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOB aiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGa GaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsga caqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabkhacaqGUaGaaeiDai aab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGa ae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7caaMc8+aaSaaaeaaca WGKbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9maalaaabaGaamiz aaqaaiaadsgacaWG4baaaiaabwgadaahaaWcbeqaaiaa=Hhadaahaa adbeqaaiaa=ndaaaaaaaGcbaGaaCzcaiabg2da9iaabwgadaahaaWc beqaaiaa=Hhadaahaaadbeqaaiaa=ndaaaaaaOWaaSaaaeaacaWGKb aabaGaamizaiaadIhaaaGaamiEamaaCaaaleqabaGaaG4maaaakiaa xMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7daWadaqaaiaabkeacaqG5bGaaeiiaiaaboeacaqGObGa aeyyaiaabMgacaqGUbGaaeiiaiaabkfacaqG1bGaaeiBaiaabwgaai aawUfacaGLDbaaaeaacaWLjaGaeyypa0JaaeyzamaaCaaaleqabaGa a8hEamaaCaaameqabaGaa83maaaaaaGccqGHxdaTcaaIZaGaamiEam aaCaaaleqabaGaaGOmaaaaaOqaaiaaxMaacqGH9aqpcaaIZaGaamiE amaaCaaaleqabaGaaGOmaaaakiaab6cacaqGLbWaaWbaaSqabeaaca WF4bWaaWbaaWqabeaacaWFZaaaaaaaaaaa@AC6E@$

Q.4

$\mathbf{\text{Differentiate}}\text{}\mathbf{\text{the}}\text{}\mathbf{\text{following}}\text{}\mathbf{\text{w}}\text{.}\mathbf{\text{r}}\text{.}\mathbf{\text{t}}\text{.}\mathbf{\text{x}}\text{: sin}{\text{(tan}}^{–1}{\mathrm{e}}^{–\mathrm{x}}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\mathrm{sin}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sin}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)\\ =\mathrm{cos}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\left[\mathrm{By}\mathrm{Chain}\mathrm{Rule}\right]\\ =\mathrm{cos}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)×\frac{1}{1+{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{2}}×\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{-\mathrm{x}}\\ =\frac{\mathrm{cos}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)}{1+{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{2}}×{\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(-\mathrm{x}\right)\\ =\frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{cos}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)}{1+{\left({\mathrm{e}}^{-\mathrm{x}}\right)}^{2}}×\left(-1\right)\\ =-\frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{cos}\left({\mathrm{tan}}^{-1}{\mathrm{e}}^{-\mathrm{x}}\right)}{1+{\mathrm{e}}^{-2\mathrm{x}}}\end{array}$

Q.5

$\mathbf{\text{Differentiate}}\text{}\mathbf{\text{the}}\text{}\mathbf{\text{following}}\text{}\mathbf{\text{w}}\text{.}\mathbf{\text{r}}\text{.}\mathbf{\text{t}}\text{.}\mathbf{\text{x}}\text{: log}{\text{(cos\hspace{0.17em}e}}^{\mathrm{x}}\right)$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\mathrm{log}\left({\mathrm{cose}}^{\mathrm{x}}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left({\mathrm{cose}}^{\mathrm{x}}\right)\\ =\frac{1}{{\mathrm{cose}}^{\mathrm{x}}}×\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{cose}}^{\mathrm{x}}\left[\mathrm{By}\mathrm{Chain}\mathrm{Rule}\right]\\ =\frac{1}{{\mathrm{cose}}^{\mathrm{x}}}×-{\mathrm{sine}}^{\mathrm{x}}×\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}\\ =\frac{-{\mathrm{sine}}^{\mathrm{x}}}{{\mathrm{cose}}^{\mathrm{x}}}×{\mathrm{e}}^{\mathrm{x}}\\ =\frac{-{\mathrm{e}}^{\mathrm{x}}{\mathrm{sine}}^{\mathrm{x}}}{{\mathrm{cose}}^{\mathrm{x}}}\\ =-{\mathrm{e}}^{\mathrm{x}}{\mathrm{tane}}^{\mathrm{x}}, {\mathrm{e}}^{\mathrm{x}}\ne \left(2\mathrm{n}+1\right)\frac{\mathrm{\pi }}{2},\mathrm{ }\mathrm{n}\in \mathrm{N}\end{array}$

Q.6

$\begin{array}{l}\mathbf{Differentiate}\mathrm{}\mathbf{the}\mathrm{}\mathbf{following}\mathrm{}\mathbf{w}.\mathbf{r}.\mathbf{t}.\mathrm{}\mathbf{x}:\\ {\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{{\mathrm{x}}^{2}}+...+{\mathrm{e}}^{{\mathrm{x}}^{5}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}...+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{5}}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}...+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{5}}}\right)\\ =\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{\mathrm{x}}+\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{2}}}+\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{3}}}+\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{4}}}+\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{5}}}\\ ={\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{2}}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{2}\right)+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{3}}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{3}\right)+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{4}}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{4}\right)+{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{5}}}\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{5}\right)\\ \left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right]\\ ={\mathrm{e}}^{\mathrm{x}}+2{\mathrm{xe}}^{{\mathrm{x}}^{\mathrm{2}}}+3{\mathrm{x}}^{2}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{3}}}+4{\mathrm{x}}^{3}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{4}}}+5{\mathrm{x}}^{4}{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{5}}}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Differentiate}\mathrm{}\mathrm{the}\mathrm{}\mathrm{following}\mathrm{}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{}\mathrm{x}:\\ \sqrt{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}, \mathrm{x}>0\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\sqrt{{\mathrm{e}}^{\sqrt{\mathrm{x}}}},\mathrm{ }\mathrm{x}>0\\ \mathrm{then},\\ {\mathrm{y}}^{2}={\mathrm{e}}^{\sqrt{\mathrm{x}}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{y}}^{2}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{\sqrt{\mathrm{x}}}\right)\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{\sqrt{\mathrm{x}}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\mathrm{x}}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}{2\mathrm{y}}.\frac{1}{2\sqrt{\mathrm{x}}}\\ \mathrm{ }=\frac{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}{4\sqrt{\mathrm{x}}\sqrt{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}}\\ \therefore \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}{4\sqrt{{\mathrm{xe}}^{\sqrt{\mathrm{x}}}}}, \mathrm{x}>0\end{array}$

Q.8

$\begin{array}{l}\mathrm{Differentiate}\mathrm{}\mathrm{the}\mathrm{}\mathrm{following}\mathrm{}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{}\mathrm{x}:\\ \mathrm{log}\left(\mathrm{logx}\right),\mathrm{ }\mathrm{x}>1\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\mathrm{log}\left(\mathrm{logx}\right),\mathrm{x}>1\\ \mathrm{Differentiating}g\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\mathrm{log}\left(\mathrm{logx}\right)\right\}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{logx}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{logx}}.\frac{1}{\mathrm{x}}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{xlogx}}, \mathrm{ }\mathrm{x}>1\end{array}$

Q.9

$\begin{array}{l}\mathbf{Differentiate}\mathrm{}\mathbf{the}\mathrm{}\mathbf{following}\mathrm{}\mathbf{w}.\mathbf{r}.\mathbf{t}.\mathrm{}\mathbf{x}:\\ \frac{\mathrm{cosx}}{\mathrm{logx}},\mathrm{ }\mathbf{x}>\mathbf{0}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{y}=\frac{\mathrm{cosx}}{\mathrm{logx}},\mathrm{x}>0\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{cosx}}{\mathrm{logx}}\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{logx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}-\mathrm{cosx}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{logx}}{{\left(\mathrm{logx}\right)}^{2}}\left[\mathrm{By} \mathrm{Quotient}\mathrm{ }\mathrm{Rule}\right]\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{logx}×-\mathrm{sinx}-\mathrm{cosx}×\frac{1}{\mathrm{x}}}{{\left(\mathrm{logx}\right)}^{2}}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\left(\mathrm{xlogx}.\mathrm{sinx}+\mathrm{cosx}\right)}{\mathrm{x}{\left(\mathrm{logx}\right)}^{2}}, \mathrm{ }\mathrm{x}>0\end{array}$

Q.10

$\begin{array}{l}Differentiate\text{}the\text{}following\text{}w.r.t.\text{}x:\\ \text{}\text{}cos\left(logx+{e}^{x}\right),\text{\hspace{0.17em}}x>0\end{array}$

Ans

$\begin{array}{l}Let\text{y}=\mathrm{cos}\left(\mathrm{log}x+{e}^{x}\right),x>0\\ Differentiating\text{both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{d}{dx}y=\frac{d}{dx}\mathrm{cos}\left(\mathrm{log}x+{e}^{x}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=-\mathrm{sin}\left(\mathrm{log}x+{e}^{x}\right)\frac{d}{dx}\left(\mathrm{log}x+{e}^{x}\right)\text{}\text{}\left[\text{Using}\text{\hspace{0.17em}}chain\text{rule}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=-\mathrm{sin}\left(\mathrm{log}x+{e}^{x}\right)\left(\frac{1}{x}+{e}^{x}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{dy}{dx}=-\left(\frac{1}{x}+{e}^{x}\right).\mathrm{sin}\left(\mathrm{log}x+{e}^{x}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0\end{array}$