Question

Differentiate w.r.t. x the function:\(x^x+x^a+a^x+a^a\), for some fixed \(a>0\) and \(x>0\)

Answer 1

The correct answer is \(=x^x(1+logx)+ax^{a-1}+a^xloga\)
Let \(y=x^x+x^a+a^x+a^a\)
Also,let \(x^x=u,x^a=v,a^x=w\) and \(a^a=s\)
\(∴y=u+v+w+s\)
\(⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}+\frac{ds}{dx}    ......(1)\)
\(u=x^x\)
\(⇒logu=logx^x\)
\(⇒logu=xlogx\)
Differentiating both sides with respect to \(x\),we obtain
\(\frac{1}{u}\frac{du}{dx}=logx.\frac{d}{dx}(x)+x.\frac{d}{dx}(logx)\)
\(⇒\frac{du}{dx}=u[logx.1+x.\frac{1}{x}]\)
\(⇒\frac{du}{dx}=x^x[logx+1]=x^x(1+logx)     .....(2)\)
\(v=x^a\)
\(∴\frac{dv}{dx}=\frac{d}{dx}(x^n)\)
\(⇒\frac{dv}{dx}=ax^{a-1}     ......(3)\)
\(w=a^x\)
\(⇒logw=loga^x\)
\(⇒logw=xloga\)
Differentiating both sides with respect to \(x\),we obtain
\(\frac{1}{w}.\frac{dw}{dx}=loga.\frac{d}{dx}(x)\)
\(⇒\frac{dw}{dx}=w\,loga\)
\(⇒\frac{dw}{dx}=a^xloga    ......(4)\)
\(s=a^a \)
Since \(a\) is constant,\(a^a\) is also a constant.
\(∴\frac{ds}{dx}=0   ......(5)\)
From (1),(2),(3),(4),and (5),we obtain
\(\frac{dy}{dx}=x^x(1+logx)+ax^{a-1}+a^xloga+0\)
\(=x^x(1+logx)+ax^{a-1}+a^xloga\)