Question

Differentiate w.r.t. x the function:\(x^{x^2-3}+(x-3)^{x^2},for\, x>3\)

Answer 1

The correct answer is: \(x^{x^2-3}.[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]\)
Let \(y=x^{x^2-3}+(x-3)^{x^2}\)
Also,let \(u=x^{x^2-3}\) and \(v=(x-3)^{x^2}\)
\(∴y=u+v\)
Differentiating both sides with respect to \(x\),we obtain
\(\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}   .....(1)\)
\(u=x^{x^2-3}\)
\(∴logu=log(x^{x^2-3})\)
\(logu=(x^2-3)logx\)
Differentiating with respect to \(x\),we obtain
\(\frac{1}{u}.\frac{du}{dx}=logx.\frac{d}{dx}(x^2-3)+(x^2-3).\frac{d}{dx}(logx)\)
\(⇒\frac{1}{u}\frac{du}{dx}=logx.2x+(x^2-3).\frac{1}{x}\)
\(⇒\frac{du}{dx}=x^{x^2-3}.[\frac{x^2-3}{x}+2x\,logx]\)
Also,
\(v=(x-3)^{x^2}\)
\(∴logv=log(x-3)^{x^2}\)
\(⇒logv=x^2log(x-3)\)
Differentiating both sides with respect to \(x\),we obtain
\(\frac{1}{v}.\frac{dv}{dx}=log(x-3).\frac{d}{dx}(x^2)+x^2.\frac{d}{dx}[log(x-3)]\)
\(⇒\frac{1}{v}\frac{dv}{dx}=log(x-3).2x+x^2.\frac{1}{x}-3.\frac{d}{dx}(x-3)\)
\(⇒\frac{dv}{dx}=v[2xlog(x-3)+\frac{x^2}{x-3}.1]\)
\(⇒\frac{dv}{dx}=(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]\)
Substituting the expressions of \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) in equation (1), we obtain
\(\frac{dy}{dx}=x^{x^2-3}.[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]\)