Find d y d x if, : y=10^ x^x +10^ x^ 10 +10^ 10^x - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$
y=10^{x^x}+10^{x^{10}}+10^{10^x}
$

✓

Answer

$
y=10^{x^x}+10^{x^{10}}+10^{10^x}
$
Let $u=x^x$
Then $\log u=\log x^x=x \log x$ Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{u} \cdot \frac{d u}{d x} & =\frac{d}{d x}(x \log x) \\
& =x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{x}+(\log x) \times 1 \\
\therefore \frac{d u}{d x} & =u(1+\log x)=x^x(1+\log x)
\end{aligned}
$
Now, $y=10^u+10^{x^{10}}+10^{10^x}$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(10^u\right)+\frac{d}{d x}\left(10^{x^{10}}\right)+\frac{d}{d x}\left(10^{10^x}\right) \\
& =10^u \cdot \log 10 \cdot \frac{d u}{d x}+10^{x^{10}} \cdot \log 10 \cdot \frac{d}{d x}\left(x^{10}\right)+ \\
\end{aligned}
$
$
\begin{array}{r}
=10^{x^x} \cdot \log 10 \cdot x^x(1+\log x)+10^{x^{10}} \cdot \log 10 \times 10^{x^9}+ \\
10^{10^x} \cdot \log 10 \times 10^x \cdot \log 10 \ldots[\text { By (1)] }
\end{array}
$
$
\begin{aligned}
& \therefore \frac{d y}{d x}=10^{x^x} \cdot x^x \cdot(\log 10)(1+\log x)+ \\
& \quad 10^{x^{10}}\left(10^{x^9}\right) \log 10+10^{10^x} \cdot 10^x \cdot(\log 10)^2 .
\end{aligned}
$

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