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

Question

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

✓

Answer

$
x =\sqrt{1+u^2}, y =\log \left(1+ u ^2\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(\sqrt{1+u^2}\right)=\frac{1}{2 \sqrt{1+u^2}} \cdot \frac{d}{d u}\left(1+u^2\right) \\
= & \frac{1}{2 \sqrt{1+u^2}} \times(0+2 u)=\frac{u}{\sqrt{1+u^2}}
\end{aligned} \\
& \text { and } \frac{d y}{d u}=\frac{d}{d u}\left[\log \left(1+u^2\right)\right] \\
& =\frac{1}{1+u^2} \cdot \frac{d}{d u}\left(1+u^2\right) \\
& =\frac{1}{1+u^2} \times(0+2 u)=\frac{2 u}{1+u^2} \\
& \begin{aligned}
\therefore \frac{d y}{d x}= & \frac{(d y / d u)}{(d x / d u)}=\frac{\left(\frac{2 u}{1+u^2}\right)}{\left(\frac{u}{\sqrt{1+u^2}}\right)} \\
& =\frac{2 u}{1+u^2} \times \frac{\sqrt{1+u^2}}{u}=\frac{2}{\sqrt{1+u^2}}
\end{aligned}
\end{aligned}
$

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