Question

The derivative of $y=x^{2^x}$ w.r.t x is :

• A. $x^{2^x}{2}^x\left(\frac{1}{x}+\ln x\ln 2\right)$
• B. $x^{2^x}\left(\frac{1}{x}\ln x\ln 2\right)$
• C. $x^{2^x}{2}^x\left(\frac{1}{x}\ln x\right)$
• D. $x^{2^x}{2}^x\left(\frac{1}{x}+\frac{\ln x}{\ln 2}\right)$

Answer 1

The correct option is A $x^{2^x}{2}^x\left(\frac{1}{x}+\ln x\ln 2\right)$

We have, $y=x^{2^x}$

take logarithm on the both sides we get,

$\ln y=2^x\ln x$

diff. w.r.t. $x$

$\frac{1}{y}\frac{dy}{dx}=[2^x\log 2.\ln x+\frac{2^x}{x}]$

$\frac{dy}{dx}=2^x.x^{2^x}[\ln 2.\ln x+\frac{1}{x}]$

$=2^x.x^{2^x}[\frac{1}{x}+\ln x\ln 2]$