Question

If $y=e^{2^x}-\cos \left(\ln x\right)$, then $\frac{dy}{dx}$ is

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

Answer 1

The correct option is B $2^x{e}^{2^x}\ln 2+\frac{\sin \left(\ln x\right)}{x}$

Given, $y=e^{2^x}-\cos \left(\ln x\right)$
So, we have using chain rule,
$\frac{dy}{dx}=\frac{d(e^{2^x}-\cos (\ln x\left)\right)}{dx}$
$\Rightarrow \frac{d{e}^{2^x}}{d{2}^x}\times \frac{d{2}^x}{dx}-\frac{d\cos \left(\ln x\right)}{d\ln x}\times \frac{d\ln x}{dx}$
$\Rightarrow 2^x{e}^{2^x}\ln 2+\frac{\sin \left(\ln x\right)}{x}$