Mathematics · 2022
JEE Main · 27 June 2022, Shift 2 · Q26
If y(x)=(x^x)^x, x>0, then (d^2 x)/(d y^2)+20 at x=1 is equal to \ \ \ \.
If $\displaystyle y(x)=\left(x^x\right)^x, x>0$, then $\displaystyle \frac{d^2 x}{d y^2}+20$ at $\displaystyle x=1$ is equal to \_\_\_\_.
Official answer
16
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.