Q106 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
$\frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]=$
- Correct answerA$x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]+(\sin x)^x$ $$ $x^{\sin x}\left[\frac{\sin x}{x}+\sin x \log x\right]+(\sin x)^x[\log \cos x+x \cot x]$ $$
- B$x^{\sin x}[x \tan x+\cos x \log x]+(\sin x)^x$ $$ \left[\frac{\sin x}{x}+\log (\sin x)\right] $$
- C$x^{\sin x}\left[\frac{x}{\sin x}+\cos x \log x\right]+(\sin x)^x$ $$ [x \cot x+\log (\sin x)] $$
- D$x^{\sin x}\left[\frac{\sin x}{x}+\sin x \log x\right]+(\sin x)^x$ $$ \text { [ } x \cot x+\log (\cos x)] $$
Solution
Given, $\frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]$
Let $U=x^{\sin x}, V=(\sin x)^x$
$$
\begin{aligned}
\log U & =\sin x \log x \\
\frac{1}{U} \cdot \frac{d U}{d x} & =\frac{\sin x}{x}+\cos x \log x \\
\frac{d U}{d x} & =x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right]
\end{aligned}
$$
and
$$
V=(\sin x)^x
$$
$$
\begin{aligned}
& \log V=x \log \sin x \\
& \frac{1}{V} \cdot \frac{d V}{d x}=\log \sin x+x \cdot \frac{1}{\sin x} \cos x \\
& \frac{d V}{d x}=(\sin x)^x[\log \sin x+x \cot x] \\
& \Rightarrow \quad \frac{d}{d x}\left[x^{\sin x}+(\sin x)^x\right]=\frac{d U}{d x}+\frac{d V}{d x} \\
&=x^{\sin x}\left[\frac{\sin x}{x}+\cos x \log x\right] \\
&+(\sin x)^x[\log \sin x+x \cot x] .
\end{aligned}
$$
Hence, the correct answer is option: 'A'.