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

Question

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

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Answer

$x=\sqrt{1+u^2}$
Differentiating both sides w.r.t. u, we get
$ \frac{ d x}{ du }=\frac{ d }{ du }\left(\sqrt{1+ u ^2}\right)$
$=\frac{1}{2 \sqrt{1+ u ^2}} \cdot \frac{ d }{ du }\left(1+ u ^2\right)$
$=\frac{1}{2 \sqrt{1+ u ^2}} \times(0+2 u )$
$=\frac{ u }{\sqrt{1+ u ^2}}$
$y=\log \left(1+ u ^2\right) $
Differentiating both sides w.r.t. u, we get
$\frac{ d y}{ du }=\frac{ d }{ du }\left[\log \left(1+ u ^2\right)\right]$
$=\frac{1}{1+ u ^2} \cdot \frac{ d }{ du }\left(1+ u ^2\right)$
$=\frac{1}{1+ u ^2} \times(0+2 u )=\frac{2 u }{1+ u ^2}$
$\therefore \frac{ d y}{ d x}=\frac{\left(\frac{ d y}{ du }\right)}{\left(\frac{ d x}{ du }\right)}=\frac{\left(\frac{2 u }{1+ u ^2}\right)}{\left(\frac{ u }{\sqrt{1+ u ^2}}\right)}$
$=\frac{2}{1+ u ^2} \times \sqrt{1+ u ^2}$
$\therefore \frac{ d y}{ d x}=\frac{2}{\sqrt{1+ u ^2}}$

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