# differentiation-chain rule- function of two variables

Problem

Let $f\left ( x,y \right )$ be a differentiable function on $\mathbb{R}^{2}$.

Find a formula for $\frac{d}{dt}f\left ( t,t^{2} \right )\mid _{t=1}$ in terms of the partial derivatives: $\frac{\partial f}{\partial x}\left ( 1,1 \right )$ and $\frac{\partial f}{\partial y}\left ( 1,1 \right )$

I started to solve the problem as follows: I applied the chain rule, and then I got: $$\frac{d}{dt}f\left ( t,t^{2} \right )=f_{1}\left ( t,t^{2} \right )\cdot 1+f_{2}\left ( t,t^{2} \right )\cdot 2t$$ and as $t\rightarrow 1$, I get: $$\frac{d}{dt}f\left ( t,t^{2} \right )\Big| _{t=1}=\frac{\partial f}{\partial x}\left ( 1,1 \right )+2\frac{\partial f}{\partial y}\left ( 1,1 \right ).$$

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Yes, your calculation of $\frac{d}{dt} f(t,t^2) \big|_{t=1}$ is correct. But there is a typo in the question (I'm not saying you made it: perhaps it is made in wherever you saw the problem).

My guess is that either

• the references to the partial derivatives of $f$ at $(0,0)$ were supposed to be references to partial derivatives of $f$ at $(1,1)$, or

• the point of evaluation $t=1$ was supposed to be $t=0$.

Because, of course, as your general calculation shows, for any real $a$ the value of $\frac{d}{dt} f(t,t^2) \big|_{t=a}$ depends only on the values of the partial derivatives of $f$ at $(a,a^2)$, and not on the values of these functions at any other point.

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Thanks for pointing out this to me. Yes, the original statement meant the partial derivatives at (1,1). I corrected the problem statement now. –  M.Krov Feb 14 '12 at 16:59