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I have $$z= f(x-ay) + g(x+ay)$$ and I have to prove that $$\frac{\partial^2 z}{\partial y^2} = a^{2} \frac{\partial^2 z}{\partial x^2}$$ but I can't understand the way partial derivative works here.

Could anyone make it a little more clear to me?

Alex Silva
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1 Answers1

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For example, $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}f(x-ay) + \frac{\partial}{\partial y}g(x+ay) =-af'(x-ay) +ag'(x+ay),$ where all I have used is the chain rule.

Thus, taking the derivative with respect to y again, $\frac{\partial^2 z}{\partial y^2} = a^2f''(x-ay) +a^2g''(x+ay)$.

Similarly, show $\frac{\partial^2 z}{\partial x^2} = f''(x-ay) +g''(x+ay)$ to show that $\frac{\partial^2 z}{\partial y^2} = a^{2} \frac{\partial^2 z}{\partial x^2}.$

Alex Silva
  • 3,557