Question Number 112720 by deepak@7237 last updated on 09/Sep/20
$$\boldsymbol{\mathrm{Q}}.\:\boldsymbol{{y}}\:=\:\left(\boldsymbol{{x}\mathrm{cos}{x}}\right)^{\boldsymbol{{x}}} +\:\left(\boldsymbol{{x}\mathrm{sin}{x}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} \:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}} \\ $$
Answered by deepak@7237 last updated on 09/Sep/20
Answered by 1549442205PVT last updated on 09/Sep/20
![Put f(x)=(xcosx)^x ,g(x)=(xsinx)^(1/x) ln[f(x)]=xln(xcosx),ln[g(x)]=(1/x)ln(xsinx) Derivative two sides of each equality by x we get: ((f ′(x))/(f(x)))=ln(xcosx)+x.[(1/(xcosx))×(cosx−xsinx)] =ln(xcosx)+((cosx−xsinx)/(cosx)) ⇒f ′(x)=(xcosx)^x ×[ln(xcosx)+((cosx−xsinx)/(cosx))] ((g′(x))/(g(x)))=(−(1/x^2 ))ln(xsinx)+(1/x)[(1/(xsinx))×(sinx+xcosx)] =−(1/x^2 )ln(xsinx)+((sinx+xcosx)/(x^2 sinx)) ⇒g′(x)=(xsinx)^(1/x) ×[(1/x^2 )ln(xsinx)+((sinx+xcosx)/(x^2 sinx))] (dy/dx)=f ′(x)+g′(x) =(xcosx)^x ×[ln(xcosx)+((cosx−xsinx)/(cosx))] +(xsinx)^(1/x) ×[(1/x^2 )ln(xsinx)+((sinx+xcosx)/(x^2 sinx))]](https://www.tinkutara.com/question/Q112742.png)
$$\mathrm{Put}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{xcosx}\right)^{\mathrm{x}} ,\mathrm{g}\left(\mathrm{x}\right)=\left(\mathrm{xsinx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\mathrm{ln}\left[\mathrm{f}\left(\mathrm{x}\right)\right]=\mathrm{xln}\left(\mathrm{xcosx}\right),\mathrm{ln}\left[\mathrm{g}\left(\mathrm{x}\right)\right]=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{xsinx}\right) \\ $$$$\mathrm{Derivative}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{each}\:\mathrm{equality} \\ $$$$\mathrm{by}\:\mathrm{x}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\frac{\mathrm{f}\:'\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\mathrm{ln}\left(\mathrm{xcosx}\right)+\mathrm{x}.\left[\frac{\mathrm{1}}{\mathrm{xcosx}}×\left(\mathrm{cosx}−\mathrm{xsinx}\right)\right] \\ $$$$=\mathrm{ln}\left(\mathrm{xcosx}\right)+\frac{\mathrm{cosx}−\mathrm{xsinx}}{\mathrm{cosx}} \\ $$$$\Rightarrow\mathrm{f}\:'\left(\mathrm{x}\right)=\left(\mathrm{xcosx}\right)^{\mathrm{x}} ×\left[\mathrm{ln}\left(\mathrm{xcosx}\right)+\frac{\mathrm{cosx}−\mathrm{xsinx}}{\mathrm{cosx}}\right] \\ $$$$\frac{\mathrm{g}'\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\left(−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{ln}\left(\mathrm{xsinx}\right)+\frac{\mathrm{1}}{\mathrm{x}}\left[\frac{\mathrm{1}}{\mathrm{xsinx}}×\left(\mathrm{sinx}+\mathrm{xcosx}\right)\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{xsinx}\right)+\frac{\mathrm{sinx}+\mathrm{xcosx}}{\mathrm{x}^{\mathrm{2}} \mathrm{sinx}} \\ $$$$\Rightarrow\mathrm{g}'\left(\mathrm{x}\right)=\left(\mathrm{xsinx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} ×\left[\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{xsinx}\right)+\frac{\mathrm{sinx}+\mathrm{xcosx}}{\mathrm{x}^{\mathrm{2}} \mathrm{sinx}}\right] \\ $$$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{f}}\:'\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{g}}'\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$=\left(\boldsymbol{\mathrm{xcosx}}\right)^{\boldsymbol{\mathrm{x}}} ×\left[\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{xcosx}}\right)+\frac{\boldsymbol{\mathrm{cosx}}−\boldsymbol{\mathrm{xsinx}}}{\boldsymbol{\mathrm{cosx}}}\right] \\ $$$$+\left(\boldsymbol{\mathrm{xsinx}}\right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} ×\left[\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{xsinx}}\right)+\frac{\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{xcosx}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{sinx}}}\right] \\ $$