If $y=\left(x^x\right)^ x$, then $\frac{d y}{d x}=$
If $y=\left(x^x\right)^ x$, then $\frac{d y}{d x}=$
- $x^{x^2}(1+\log x)$
- $x \cdot x^{x^2}(1+\log x)$
- $x^{x^2}(1+2 \log x)$
- $x \cdot x^{x^2}(2 \log x+1)$
Solution
$y=\left(x^x\right)^x \Rightarrow \log y=x \log x^x=x^2 \log x$
Differentiating both sides w.r.t. $\mathrm{x}$
$\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=x^2 \times \frac{1}{x}+2 x \times \log x \\ & \Rightarrow \frac{d y}{d x}=y(x+2 x \log x) \\ & \Rightarrow \frac{d y}{d x}=\left(x^x\right)^x \cdot x(1+2 \log x) \\ & \Rightarrow \frac{d y}{d x}=x^{x^2} \cdot x(1+2 \log x)\end{aligned}$
Asked in: MHT CET 2022 (05 Aug Shift 1)
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