Q99 / 180 · KCET 2021 · Mathematics
For constant $a, \frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is
- A$x^{x}(1+\log x)+a x^{a-1}$
- Correct answerB$x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$
- C$x^{x}(1+\log x)+a^{a}(1+\log x)$
- D$x^{x}(1+\log x)+a^{a}(1+\log a)+a x^{a-1}$
Solution
$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$
$\begin{aligned}
&=\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right) \\
&=\frac{d}{d x}(y)+a x^{a-1}+a^{x} \log a+0...(i)
\end{aligned}$
$\left[\operatorname{let} x^{x}=y\right]$
$y=x^{x}$
Taking log on both sides,
$\log y=x \log x$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{1}{y} \frac{d y}{d x} &=x \times \frac{1}{x}+\log x \\
\Rightarrow \quad \quad \frac{d y}{d x} &=x^{x}(1+\log x)
\end{aligned}$
Substitute the value of $\frac{d y}{d x}$ in Eq. (i),
$\begin{aligned}
&\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right) \\
&=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a
\end{aligned}$
Hence, the correct answer is option: 'B'.