If I take partial derivatives of function using chain rule, how do I remove variable from the partials in the equation?

Question

I have the function $$f(tx,ty)$$ and I want to take the partial derivative of this with respect to $t$. So set $x'=xt$ and $y'=yt$. I applied chain rule and got

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x'}\frac{\partial x'}{\partial t} +\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial t}$$

which yields

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x'}x +\frac{\partial f}{\partial y'}y$$

But how do I evaluate that? It is still in terms of $x'$ and $y'$. How do I get it in terms of partials of $x$ and $y$?

Answer 1

I find your notation a little confusing so I'll restate this with the notation I'm used to:

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$

Like you say, this is still in terms of $x$ and $y$ at least in some of it, like $\frac{\partial f}{\partial x}$. But $x$ is a function of $t$ so you can substitute $x$ for that function, and likewise for $y$.

For instance, if $f(x,y)=xy^2$ and $x=\sin t$ and $y=e^t$ then

$$\frac{\partial f}{\partial t} = (y^2)(\cos t) + 2xy(e^t)$$

$$=e^{2t}\cos t+ 2(\sin t)(e^{t})(e^t)$$