If $y=x^{ an^{-1}x}$, then $\dfrac{dy}{dx}$ equals

Question

If $y=x^{	an^{-1}x}$, then $\dfrac{dy}{dx}$ equals

Accepted Answer

A. $x^{	an^{-1}x}\left(\dfrac{	an^{-1}x}{x}+\dfrac{\log x}{1+x^2}ight)$

For a variable base and variable exponent, logarithmic differentiation is convenient. Write $\log y=	an^{-1}x\cdot \log x$. Differentiating, $\frac{1}{y}\frac{dy}{dx}=\frac{	an^{-1}x}{x}+\frac{\log x}{1+x^2}$. Multiplying by $y=x^{	an^{-1}x}$ gives option A.

Answer

$x^{	an^{-1}x}\left(\dfrac{1}{x}+\dfrac{1}{1+x^2}ight)$

Answer

$x^{	an^{-1}x}\left(	an^{-1}x+\log xight)$

Answer

$x^{	an^{-1}x}\left(\dfrac{	an^{-1}x}{x}-\dfrac{\log x}{1+x^2}ight)$

If $y=x^{\tan^{-1}x}$, then $\dfrac{dy}{dx}$ equals