MCQ
If $y ={x^{{x^2}}}$ then $\frac{{dy}}{{dx}}=$
- A$2 \, ln \, x .{x^{{x^2}}}$
- B$(2 \, ln \, x + 1).{x^{{x^2}}}$
- C$(2 \, ln \, x + 1).{x^{{x^2+2}}}$
- ✓${x^{{x^2}}}. \, ln \, ex^2$
Taking log on both the sides, we get $\log y=x \log x$
On differentiating w.r.t. $x$, we get $\frac{1}{y} \frac{d y}{d x}=\frac{x}{x}+\log x$
$\Rightarrow \frac{d y}{d x}=y+y \log x$
$\Rightarrow \frac{d y}{d x}=x^{x}(1+\log x) \ldots\left(\because y=x^{x}\right)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.