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If there is a function $z=f(x^2 + y^2)$

What is the partial differential of $z$ w.r.t to ?

I think it is equal to the partial differential f w.r.t to $x$ multiplied by $2x$.

But the answer is wrong. It includes some total derivative which I don't understand .

First of all I want to know why is the above wrong ? What is wrong in the above chain rule application? Why do we have use the total derivative ?

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  • $\begingroup$ I think you wanted to say ”What is the partial differential of z w.r.t to x ' $\endgroup$
    – Shashaank
    Commented Dec 28, 2016 at 18:51

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You have a typo in your question - what's the partial derivative of $z$ with restpect to what? However I think I might have a hint for you nevertheless. Notice how $f$ is a function of really just one variable (while $z = z(x, y)$ is a function of two variables), so there is no partial derivative of $f$ per se. This may be confusing, but look at $x^2 + y^2$ as a whole: $$ z = f(\xi), $$ where $z = z(x, y)$ and $\xi = \xi(x, y)$, particularly $\xi(x, y) = x^2 + y^2$. Then $$ \frac{\partial z}{\partial x} = \frac{\mathrm{d} f}{\mathrm{d} \xi} \cdot \frac{\partial \xi}{\partial x} $$ or simply $$ \frac{\partial z}{\partial x} = f'(x^2 + y^2) \cdot 2x $$

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