Question

Differentiate the function with respect to \(x\):
\(x^{sin\,x}+(sin\,x)^{cos\,x}\)

Answer 1

The correct answer is \(\frac{dy}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}]+(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx]\)
Let \(y=x^{sin\,x}+(sin\,x)^{cosx}\)
Also,let \(u=x^{sinx}\) and \(v=(sinx)^{cosx}\)
\(∴y=u+v\)
\(⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}   ....(1)\)
\(u=x^{sinx}\)
\(⇒log\,u=log(x^{sin\,x})\)
\(⇒log\,u=sin\,x\,log\,x\)
Differentiating both sides with respect to \(x\),we obtain 
\(⇒\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(sin\,x)log(x)+sinx.\frac{d}{dx}[log\,x]\)
\(⇒\frac{du}{dx}=u[cos\,x\,log\,x+sinx.\frac{1}{x}]\)
\(⇒\frac{du}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}]    ..(2)\)
\(v=(sin\,x)^{cosx}\)
\(⇒log\,v=log(sin\,x)^{cosx}\)
\(⇒log\,v=cos\,x\,log(sin\,x)\)
Differentiating both sides with respect to \(x\), we obtain
\(\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(cosx)\times log(sinx)+cosx\frac{d}{dx}(log(sinx))\)
\(⇒\frac{dv}{dx}=v[-sin\,x\,log(sinx)+cosx.\frac{1}{sinx}.\frac{d}{dx}(sinx)]\)
\(⇒\frac{dv}{dx}=(sin\,x)^{cos\,x}[-sin\,x\,log\,sin\,x+\frac{cos\,x}{sin\,x}cos\,x]\)
\(⇒\frac{dv}{dx}=(sinx)^{cosx}[-sinx\,log\,sinx+cotx\,cosx]\)
\(⇒\frac{dv}{dx}=(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx]  ...(3)\)
Therefore,from (1),(2),and(3),we obtain
\(\frac{dy}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}]+(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx]\)