If y=x^x+x^a+a^x+a^a then find d y d x . - Vidyadip

Question

If $y=x^x+x^a+a^x+a^a$ then find $\frac{d y}{d x}$.

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Answer

Suppose that
$y=x^x+x^a+a^x+a^a$
differentiating w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a+0$
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a$
Suppose that $u=x^x$
$ \therefore \log u =\log _e x^x=x \log _e x$
$\therefore \frac{1}{u} \frac{d u}{d x} =1 \cdot \log _e x+x \cdot \frac{1}{x}=1+\log _e x$
$ \frac{d}{d x}\left(x^x\right) =\frac{d u}{d x}=u\left(1+\log _e x\right)=x^x\left(1+\log _e x\right)$
Put the value of $\frac{d}{d x}\left(x^x\right)$ in equation (1)$
\frac{d y}{d x}=x^x\left(1+\log _e x\right)+a x^{a-1}+a^x \log _e a$

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