Question

The value of $\frac{d}{dx}{x}^{2x}=$

• (A) $2x^2\log x$
• (B) $\frac{x^2}{2(1+\log x)}$
• (C) $x^{2x}$
• (D) $2x^{2x}[1+\log x]$

Answer 1

The Correct Option is D

Explanation:
Given:$\frac{d}{dx}{x}^{2x}$Concept:$\log x^n=n\log x$$\frac{d}{dx}[\log x]=\frac{1}{x}$Let, $y=x^{2x}$Taking log on both the sides, we get$\log y=\log \left(x^{2x}\right)=2x\log x$$(\because \log x^n=nlogx)$Now, taking derivatives,$\frac{d}{dx}[\log y]=2\{\frac{d}{dx}[x]\log x+\frac{d}{dx}[\log x]x\}$$\frac{1}{y}\frac{dy}{dx}=2[\log x+x\cdot \frac{1}{x}]$$(\because \frac{d}{dx}[\log x]=\frac{1}{x})$$\Rightarrow \frac{dy}{dx}=2y[1+\log x]$$\Rightarrow \frac{dy}{dx}=2x^{2x}[1+\log x]$$(\because y=x^{2x})$Hence, the correct option is (D).