Find d y d x if, : x = (u+1 u )^2, y =(2)^ (u+1 u ) - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$
x =\left(u+\frac{1}{u}\right)^2, y =(2)^{\left(u+\frac{1}{u}\right)}
$

✓

Answer

$
x =\left(u+\frac{1}{u}\right)^2, y =(2)^{\left(u+\frac{1}{u}\right)}
$
Differentiating $x$ and $y$ w.r.t. $u$, we get,
$
\begin{aligned}
& \begin{aligned}
\frac{d x}{d u} & =\frac{d}{d u}\left(u+\frac{1}{u}\right)^2=2\left(u+\frac{1}{u}\right) \cdot \frac{d}{d u}\left(u+\frac{1}{u}\right) \\
& =2\left(u+\frac{1}{u}\right)\left(1-\frac{1}{u^2}\right)
\end{aligned} \\
& \text { and } \frac{d y}{d u}=\frac{d}{d u}\left[2^{\left(u+\frac{1}{u}\right)}\right]
\end{aligned}
$
$
\begin{aligned}
& =2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot \frac{d}{d u}\left(u+\frac{1}{u}\right) \\
& =2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot\left(1-\frac{1}{u^2}\right)
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x}=\frac{(d y / d u)}{(d x / d u)} & =\frac{2^{\left(u+\frac{1}{u}\right)} \cdot \log 2 \cdot\left(1-\frac{1}{u^2}\right)}{2\left(u+\frac{1}{u}\right)\left(1-\frac{1}{u^2}\right)} \\
& =\frac{2^{\left(u+\frac{1}{u}\right)} \cdot \log 2}{2\left(u+\frac{1}{u}\right)} \\
& =\frac{y \log 2}{2 \sqrt{x}}
\end{aligned}
$

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