Question

Differentiate the function with respect to $x$:
$x^x-2^{\sin x}$

• A. $\ln x+1-2^{\sin x}\ln 2\cdot \cos x$
• B. $x^x(\ln x+1)-2^{\sin x}\ln 2\cdot \cos x$
• C. $x^x(\ln x+1)-2^{\sin x}\ln 2$
• D. $x^x(\ln x+1)-2^{\sin x}\cdot \cos x$

Answer 1

Answer: (B)

$x^x(\ln x+1)-2^{\sin x}\ln 2\cdot \cos x$

Let

$y=x^x-2^{\sin x}...................\left(1\right)$

Taking \log both sides

$\log y=x\log x-\sin x\log 2$

$\frac{1}{y}\frac{dy}{dx}=x\frac{1}{x}+\log x-\cos x\log 2$

$\frac{1}{y}\frac{dy}{dx}=1+\log x-\cos x\log 2$

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

From (1)

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