Partial Differentiation | IIT JAM 2017 | Problem 5

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function. If $g(u, v)=f\left(u^{2}-v^{2}\right),$ then
$\frac{\partial^{2} g}{\partial u^{2}}+\frac{\partial^{2} g}{\partial v^{2}}=$

Here $g$ is a function of $u$ and $v$, to calculate $\frac{\partial g}{\partial u}$ we will differentiate the function $g$ with respect to $u$ keeping $v$ as constant.

and to calculate $\frac{\partial g}{\partial v}$ we will differentiate the function $g$ with respect to $v$ keeping $u$ as constant.

and $\frac{\partial^2 g}{\partial u^2}= \frac{\partial }{\partial u} [ \frac{\partial g}{\partial u} ]$

Hmm... I think you can easily do it from here ........

$\begin{aligned}\frac{\partial^{2} g}{\partial u^{2}}=\frac{\partial}{\partial u}\left(\frac{\partial u}{\partial u}\right) &=\frac{\partial}{\partial u}\left[f^{\prime}\left(v^{2}-v^{2}\right) \cdot 2 u\right] \\&=2 u \cdot f^{\prime \prime}\left(v^{2}-v^{2}\right) \cdot 2 u+f^{\prime}\left(v^{2}-v^{2}\right) \cdot 2 \\&=4 u^{2} f^{\prime \prime}\left(v^{2}-v^{2}\right)+2 f^{\prime}\left(v^{2}-v^{2}\right)\ldots\ldots(i)\end{aligned}$

Similarly,

$\begin{aligned}\frac{\partial^{2} g}{\partial v^{2}}=\frac{\partial}{\partial v}\left(\frac{\partial g}{\partial v}\right) &=\frac{\partial}{\partial v}\left[f^{\prime}\left(v^{2}-v^{2}\right) \cdot (-2 v)\right] \\&=(-2 v) \cdot f^{\prime \prime}\left(v^{2}-v^{2}\right) \cdot (-2 v)+f^{\prime}\left(v^{2}-v^{2}\right) \cdot (-2) \\&=(4 v^{2}) f^{\prime \prime}\left(v^{2}-v^{2}\right)-2 f^{\prime}\left(v^{2}-v^{2}\right) \ldots\ldots(ii) \end{aligned}$

Adding $(i)$ and (ii) we get,

$\frac{\partial^{2} g}{\partial u^{2}}+\frac{\partial^{2} g}{\partial x^{2}}$

$=4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right)+2 f^{\prime}\left(u^{2}-v^{2}\right)-2 f^{\prime}\left(u^{2}-v^{2}\right)$

$=4\left(u^{2}+v^{2}\right) f^{\prime \prime}\left(u^{2}-v^{2}\right) \textbf{[Ans]}$