Let y be an implicit function of x defined by x^2x-2x^x y-1=0.

Question

Let $y$ be an implicit function of $x$ defined by $x^{2x}-2x^x\cot y-1=0$. Then $y'(1)$ equals

Accepted Answer

Let $u=x^x$; then $2\cot y=u-\dfrac1u$. Differentiating: $-2\csc^2y\,y'=\left(1+\dfrac{1}{u^2}\right)u'$, with $u'=x^x(1+\ln x)$.

$$y'=\frac{(x^x+x^{-x})(1+\ln x)}{-2(1+\cot^2y)}$$

At $x=1$: from the original equation, $\cot y=0$. So $y'(1)=\dfrac{(1+1)(1+0)}{-2(1+0)}=-1$.

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