Question

Find $$\frac{dy}{dx}$$  $$y=x^x+{(\sin x)}^x$$ ?

Answer 1

$$\text{We have, y}=x^x+{(\sin x)}^x$$

$$\Rightarrow y=e^{\log x^x}+e^{\log {(\sin x)}^x}$$

$$\Rightarrow y=e^{x\log x}+e^{x\log \sin x}$$

Differentiating with respect to x using chain rule and product rule,

$$\frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log x}\right)+\frac{d}{dx}\left(e^{x\log \sin x}\right)$$

$$=e^{x\log x}\frac{d}{dx}(x\log x)+e^{x\log \sin x}\frac{d}{dx}(x\log \sin x)$$

$$=e^{x\log x}[x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}\left(x\right)]+e^{\log {(\sin x)}^x}[x\frac{d}{dx}(\log \sin x)+\log \sin x\frac{d}{dx}\left(x\right)]$$

$$=x^x[x\left(\frac{1}{x}\right)+\log x\left(1\right)]+{(\sin x)}^x[x\left(\frac{1}{\sin x}\right)\frac{d}{dx}(\sin x)+\log \sin x]$$

$$=x^x[1+\log x]+{(\sin x)}^x[x\left(\frac{1}{\sin x}\right)(\cos x)+\log \sin x]$$

$$=x^x[1+\log x]+{(\sin x)}^x[x\cot x+\log \sin x]$$