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If a $C^1$ function $f(u,v)$ has the following partial derivatives $$\frac{\partial f}{\partial u} (u,v) = 2uv \\ \frac{\partial f}{\partial v} (u,v) = u^2$$ and we have $g(t) =f(2t,t^2)$, what is $g'(t)$?

Trying to understand the chain rule, and it's worked out fine so far but am having problems with this one.

Git Gud
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Joaneael
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  • Let $h\colon \mathbb R\to \mathbb R^2, t\mapsto (h_1,h_2)$ where $h_1\colon \mathbb R\to \mathbb R, t\mapsto 2t$ and $h_2\colon \mathbb R\to \mathbb R, t\mapsto t^2$. You want to find $g'$ which equals $(f\circ h)'$. The chain rule tells you that $$\forall t\in \mathbb R\left[g'(t)=\dfrac{\partial f}{\partial _1}(h(t))h_1'(t)+\dfrac{\partial f}{\partial _2}(h(t))h_2'(t)\right].$$ Here $\dfrac{\partial f}{\partial _k}$ denotes the partial derivative of $f$ with respect to its $k^\text{th}$ coordinate. – Git Gud Nov 06 '14 at 10:21

1 Answers1

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Firstly, it should be straighforward enough to see that if given those partial derivatives above; then \begin{equation} f(u, v) = u^{2}v + C \end{equation} Now according to the change of variable; \begin{equation} u=2t, v=t^{2} \end{equation} We find that \begin{eqnarray} g(t) &=& f(2t, t^{2}) + C \\ &=& (2t)^{2}t^{2}+ C \\ &=& 4t^{4} + C \end{eqnarray}

And the derivative from there is straighforward. Alternatively, using the chain rule (as you specified); we can write \begin{eqnarray} \frac{dg}{dt} &=& \frac{df}{dt}(2t, t^{2}) \\ &=& \frac{\partial f}{\partial u}\frac{du}{dt}+\frac{\partial f}{\partial v} \frac{dv}{dt} \end{eqnarray} And simply fill in the blanks from there.

All the best, A

Autolatry
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