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I know that $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$ and we can then say that $$\frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx}$$ And so if $y$ is a function of $x$, then $f(x,y(x))$ $$df = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$$ My question is, if y depends on x, how can it make sense to take $\frac{\partial f}{\partial y}$ where x is being held constant? if $x$ is held constant how can $y$ vary to find the partial derivative?

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This is one of the places where the usual notation can, despite its convenience, lead to endless confusion. Almost all the symbols you've used mean different things at their different appearance: alas. And thinking of partial differentiation as "differentiating something keeping other things constant" is sometimes too naive a way to think: also, alas, because at the start it's useful.

As names of variables don't matter let's start afresh. You've got a real valued function $f(u,v)$ of two real variables. Each of these is a function of a real variable $t$, $u(t)$ and $v(t)$.

Now we look at the function $F(t)=f(u(t), v(t))$; note this is not the same as the function $f$ since this one is a function of just one real variable.

You know that by the Chain Rule, we have that $$\frac{dF}{dt} = \frac{\partial f}{\partial u}\frac{du}{dt} + \frac{\partial f}{\partial v}\frac{dv}{dt}. $$

It may help avoid some confusion if we mention explicitly the variable in each term: $$\frac{dF(t)}{dt} = \frac{\partial f(u,v)}{\partial u(t)}\frac{du}{dt} + \frac{\partial f(u,v)}{\partial v}\frac{dv(t)}{dt}. $$

Now we specialise to the case you want: $u(t)=t$, $v(t)$ still arbitrary.

$$\frac{d}{dt}F(t) = \left.\frac{\partial f(u,v)}{\partial u}\right|_{u=t,v=v(t)} + \left.\frac{\partial f(u,v)}{\partial v}\right|_{u=t,v=v(t)}\frac{dv(t)}{dt}. $$

Finally, if we wish we can remove the reference to $F$: $$\frac{d}{dt}f(t,v(t)) = \left.\frac{\partial f(u,v)}{\partial u}\right|_{u=t,v=v(t)} + \left.\frac{\partial f(u,v)}{\partial v}\right|_{u=t,v=v(t)}\frac{dv(t)}{dt}. $$

Everything is now unambiguous and these apparent contradictions vanish.

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