# Multiplying partial derivatives by normal derivatives?

I'm taking a course on multivariable calculus. The professor wrote the following:

$$f=uv, u=u(t), v=v(t)$$

$$\frac{d(uv)}{dt} = f_u\frac{du}{dt} + f_v\frac{dv}{dt}=v\frac{du}{dt}+u\frac{dv}{dt}$$

Here's the little I understand:

• $$f$$ is a function of $$u, v$$.
• $$u$$ and $$v$$ are functions of $$t$$.
• The derivative of $$f$$ with respect to $$t$$, which is written as $$\frac{d(uv)}{dt}$$, is the sum of the partial of $$f$$ with respect to $$u$$ times the derivative of $$u$$ with respect to $$t$$ and the partial of $$f$$ with respect to $$v$$ times the derivative of $$v$$ times $$t$$.

Why is this true, and what is going on in the third part of the equation (after the second equals sign)?

I guess the middle part could be written as $$\frac{\partial f}{\partial u}\frac{du}{dt} + \frac{\partial f}{\partial v}\frac{dv}{dt}$$, but how dows that simplify into $$v\frac{du}{dt}+u\frac{dv}{dt}$$?

• $\partial f/\partial u$ means to differentiate $f=uv$ treating the other variables (in this case, $v$) as constants.) Jul 18, 2014 at 19:54
• Note that this is the single variable product rule, just using some notions from multivariable calc! Jul 18, 2014 at 20:02
• @mathematician That helped a lot. Thank you Jul 18, 2014 at 22:34

Taking the partial with respect with u, or in the notation:

$$\frac{df}{du}$$

means to treat u as a variables and treat everything else as a constant.

So if $$f(u,v) = uv$$

Then $v$ is a constant and $u$ is your variable.

So when you go to differentiate:

$$\frac{df}{du}(f(u,v)) = v$$

Similarly, for the partial with respect with $v$:

$$\frac{df}{dv}(f(u,v)) = u$$

• At the last two equation there are, in each case, one f and one bracket too much. Jul 18, 2014 at 20:09
• The $d$ notation as opposed to $\partial$ will only lead to greater confusion. Could you switch? Jul 18, 2014 at 20:10
• Ah, right. Thanks. Jul 19, 2014 at 1:42

You can get the first set of terms by taking the derivative $\frac{d}{dt}$ of the total derivative of $f(u,v)$:

$df(u,v) =\frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv$

Take $\frac{d}{dt}$:

$\frac{d}{dt}df(u,v) = \frac{\partial f}{\partial u}\frac{du}{dt} + \frac{\partial f}{\partial v}\frac{dv}{dt}$

The second equality is due to the product rule as applied to $uv$:

$\frac{d(uv)}{dt} = v\frac{du}{dt}+u\frac{dv}{dt}$

Since $f=uv$ you simply equate the two to see that $v=\frac{\partial f}{\partial u}$ and $u=\frac{\partial f}{\partial v}$