Multiplying partial derivatives by normal derivatives?

Question

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 $v\frac{du}{dt}+u\frac{dv}{dt}$?

Answer 1

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 $v$:

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

Answer 2

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 $u=\frac{\partial f}{\partial v}$