# Derivative of function in tangent-plane

I have a sheet of answers for some of the assignments I had to complete, this is a step I don't really understand:

How does he go from $$f(x,y) = 5e^{x^2-y^2}$$ to $$fx(x,y) = 5e^{x^2-y^2}(-2y) = 10xe^{x^2-y^2}$$?

See source here:

This is the thing I don't understand: Say $$u$$ is 5 and $$v$$ is $$(e^{x^2-y^2})$$ than the chain rule should be applied after $$u'*v*v' = (e^{x^2-y^2}) *2x(e^{x^2-y^2})$$ (I am not sure here and unfortunately your rather formal way of doing this confuses me).

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presumably you (or the answer sheet) is missing a $\partial / \partial_y$? – Willie Wong May 24 '13 at 10:55
That is not the chain rule. That is vaguely like the product rule. Also, since $v$ is a function of $2$ variables, then $v'$ isn't really meaningful. – Cameron Buie May 26 '13 at 22:47

## 1 Answer

Now that you've posted the source, the short answer is: he doesn't.

This is simply an application of the chain rule. Put $g(t)=5e^t$, so that $f(x,y)=g(x^2-y^2).$ Then

\begin{align}f_x(x,y,) &:= \frac{\partial}{\partial x}[f(x,y)]\\ &= \frac{\partial}{\partial x}[g(x^2-y^2)]\\ &= g'(x^2-y^2)\cdot\frac{\partial}{\partial x}[x^2-y^2]\\ &= g'(x^2-y^2)\cdot 2x\\ &= g(x^2-y^2)\cdot 2x\\ &=f(x,y)\cdot 2x.\end{align} We can do something similar for $f_y$.

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These two functions, how did you calculate those? That's my very question. Did you use the product rule? – IMX May 24 '13 at 14:27
just use chain rule – PepeToro May 24 '13 at 14:49
@Cameron Buie, your way of doing this might be formally correct, but it doesn't really solve the problem for me. I edited my question once again, could you take a look? – IMX May 25 '13 at 15:32