JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1
If $y(x)=\left(x^{x}\right)^{x}$, $x>0$, then $\frac{d^{2}x}{dy^{2}}+20$ at $x=1$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$16$
Solution
Concept. $\frac{d^2x}{dy^2}=-\frac{y''}{(y')^3}$. Find $y'$ and $y''$ at $x=1$ by logarithmic differentiation. Formulas used. (1) $\dfrac{d^2x}{dy^2}=-\dfrac{y''}{(y')^3}$(2) $\ln y=x^2\ln x$ $\Rightarrow$ $y'=y(2x\ln x+x)$Step 1 - $y'$. At $x=1$: $y=1$ and by (2) $y'=1\cdot(0+1)=1$. Step 2 - $y''$. Differentiating (2): $$y''=y'(2x\ln x+x)+y(2\ln x+3).$$ At $x=1$: $y''=1+3=4$. Step 3 - Value. By (1), $\frac{d^2x}{dy^2}=-4$, so the answer is $-4+20=16$. Answer: 16