Question

If $y={\left(sinx\right)}^{cosx}+{\left(cosx\right)}^{sinx},find\frac{dy}{dx}$

Answer 1

We have,

$y={\left(\sin x\right)}^{\cos x}+{\left(\cos x\right)}^{\sin x}$

Taking log both side and we get,

$\log y=\log {\left(\sin x\right)}^{\cos x}+\log {\left(\cos x\right)}^{\sin x}$

Now,

$\log y=\cos x.\log \sin x+\sin x.\log \cos x$

On differentiating with respect to x and we get,

$\frac{d}{dx}\log y=\cos x\frac{d}{dx}\log \sin x+\log \sin x\frac{d}{dx}\cos x+\sin x\frac{d}{dx}\mathrm{logcos}x+\log \cos x\frac{d}{dx}\sin x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\cos x\frac{1}{\sin x}\cos x+\log \sin x(-\sin x)+\sin x\frac{1}{\cos x}(-\sin x)+\log \cos x\cos x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{{\cos }^{2}x}{\sin x}-sinx\log \sin x-\frac{{\sin }^{2}x}{\cos x}+\cos x\log \cos x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{{\cos }^{2}x}{\sin x}-\frac{{\sin }^{2}x}{\cos x}-sinx\log \sin x+\cos x\log \cos x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{{\cos }^{2}x}{\sin x}-\frac{{\sin }^{2}x}{\cos x}+\cos x\log \cos x-sinx\log \sin x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{{\cos }^{3}x-{\sin }^{3}x}{\sin x\cos x}+\log {\left(\cos x\right)}^{\cos x}-\log {\left(\sin x\right)}^{sinx}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{{\cos }^{3}x-{\sin }^{3}x}{\sin x\cos x}+\log \frac{{\left(\cos x\right)}^{\cos x}}{{\left(\sin x\right)}^{sinx}}$

$\Rightarrow \frac{dy}{dx}=y[\frac{{\cos }^{3}x-{\sin }^{3}x}{\sin x\cos x}+\log \frac{{\left(\cos x\right)}^{\cos x}}{{\left(\sin x\right)}^{sinx}}]$

$\Rightarrow \frac{dy}{dx}={\left(\sin x\right)}^{\cos x}+{\left(\cos x\right)}^{\sin x}[\frac{{\cos }^{3}x-{\sin }^{3}x}{\sin x\cos x}+\log \frac{{\left(\cos x\right)}^{\cos x}}{{\left(\sin x\right)}^{sinx}}]$

Hence, this is the answer.