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Suppose I have the following functions, $f(x,y)=xy-y^3$ and $g(u,v)=f(u^3,uv^2)$ then calculate the gradient $dg(u_0,v_0)$ using the chain rule.

My attempt: I set $x=u^3$ and $y=uv^2$

$\frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial g}{\partial x}(3u^2)+\frac{\partial g}{\partial y}(v^2)$

I think I understand it correctly up to this point, but how do I find $\frac{\partial g}{\partial x}$ and other terms like it?

jon
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1 Answers1

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As you set $x = u^3$, this implies $u = x^{1/3}$. Use this value of u in the expression above to find out the partial derivative of $3u^2$ w.r.t $x$.

I hope it will help you.

Adnan
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  • I'm not convinced this is the correct way to do it. Can you explain why or show that it is in fact correct? – jon Mar 10 '15 at 17:46