Q.If $x = \sqrt{1 + u^2}$, $y = \log(1 + u^2)$, then find $\dfrac{dy}{dx}$.

Differentiate $x$ and $y$ separately with respect to $u$, then divide: $\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1+u^2}}=\dfrac{2}{x}$.

Here $x=\sqrt{1+u^2}=(1+u^2)^{1/2}$ and $y=\log(1+u^2)$, both functions of the parameter $u$.

Differentiate $y$ with respect to $u$ (chain rule):

$\dfrac{dy}{du}=\dfrac{1}{1+u^2}\cdot 2u=\dfrac{2u}{1+u^2}$.

Differentiate $x$ with respect to $u$:

$\dfrac{dx}{du}=\dfrac{1}{2}(1+u^2)^{-1/2}\cdot 2u=\dfrac{u}{\sqrt{1+u^2}}$.

Now use the parametric rule:

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