Question

Evaluate:

$x^x-2^{\sin x}$.

Answer 1

given $y=x^x-2^{\sin x}$

Let $y=x^x$ and $y_2=2^{\sin x}$

then $\frac{dy}{dx}=\frac{dy}{d{x}^1}-\frac{dy}{d{x}^2}$

$\log y_1=x\log x;\log y_2=\sin x\log 2$

$\frac{1}{y_1}\frac{dy}{d{x}^1}=\log x+x\times \frac{1}{x};\frac{1}{y_2}\frac{dy}{d{x}^2}=\cos x.\log 2$

$\frac{1}{y_1}\frac{dy}{d{x}^1}=\log x+1;\frac{1}{y_2}\frac{dy}{d{x}^2}=\cos \log 2$

$\frac{dy}{d{x}^1}=y_1(1+\log x);\frac{dy}{d{x}^2}=2^{\sin x}.\cos x.\log 2$

$\frac{dy}{dx}=\frac{dy}{d{x}^1}-\frac{dy}{d{x}^2}$

$\frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}\cos x\log 2$