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. $x$
$
\begin{array}{l}
\frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a+0 \\
\Rightarrow \quad \frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a
\end{array}
$
Suppose that $u=x^x$
$\begin{array}{rlrl} \therefore \log u =\log _e x^x=x \log _e x \\ \therefore \frac{1}{u} \frac{d u}{d x} =1 \cdot \log _e x+x \cdot \frac{1}{x}=1+\log _e x \\ \frac{d}{d x}\left(x^x\right) =\frac{d u}{d x}=u\left(1+\log _e x\right)=x^x\left(1+\log _e x\right)\end{array}$
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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