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$ \mathrm{If ~u=(x^2-y^2)f(t) where~ t=xy~and~ f ~denotes~ an ~arbitrary~ function,~ prove~ that~ \frac{\partial^2{u}} {\partial{x} {y}}=(x^2-y^2){\{t\cdot{f''(t)} } +3f'(t)}\} $

My attempt:$ \frac{\partial {u} } {\partial{x} } =(2x)f(t)+f'(t) \cdot(x^2-y^2) $ Then I tried taking the $\frac{\partial^2{u}} {\partial{x}{\partial{y} }} $ but I don't seem to be getting anywhere close.

Assistance is needed on this. My knowledge is a bit elementary on partial differentiation.

Orestes Dante
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    In your $\partial_x u$ you applied the chain rule incompletely: $$\partial_x f(t) = f^\prime(t)\partial_xt = f^\prime(t)y$$ – denklo Jun 04 '19 at 07:44

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You have to use the chain rule for differentiating $f(t)$,$$\frac{\partial u}{\partial x}=2xf(t)+(x^2-y^2)\frac{df(t)}{dt}\cdot\frac{\partial t}{\partial x}=2xf(t)+\color{red}y(x^2-y^2)f'(t)$$because $\displaystyle\frac{\partial t}{\partial x}=\frac{\partial(xy)}{\partial x}=y$. Once again, keeping in mind the chain rule,$$\frac{\partial^2u}{\partial x\partial y}=\frac\partial{\partial y}[2xf(t)+y(x^2-y^2)f'(t)]\\=2x^2f'(t)+(x^2-3y^2)f'(t)+xy(x^2-y^2)f''(t)\\=(x^2-y^2)[3f'(t)+tf''(t)]$$which is the required expression.

Shubham Johri
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