Chain rule derivatives

Question

When you use the chain rule to equate $$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$$ If $\frac{dv}{dt}$ is a function of $x$, are the resulting two derivatives functions of x, as well?

Thanks!

Answer 1

Chain rule is used to compute the derivative of a composite function, say $f(g(t))$. If we let $v=f(x)$ and $x=g(t)$, the derivative of $f(g(x))$ is $v’$ which is computed using the chain rule $$ \frac{dv}{dt}= \frac{dv}{dx} \frac{dx}{dt} $$ And the resulting derivative is a function of $t$ only.

Example: Find the derivative of $v=(t^2 +t)^2$.

We can let $v=f(x)=x^2$ and $x=g(f)=t^2+t$ then $$ \frac{dv}{dx}= 2x =2(t^2+t) \quad \text{and} \quad \frac{dx}{dt}=2t+1 $$ Then, we use the chain rule $$ \frac{dv}{dt}= \frac{dv}{dx} \frac{dx}{dt} = 2(t^2+t)(2t+1) $$ Which is exactly what you’ll be getting if you differentiated $v$ directly using the power rule. You can see from here that the derivative is a function of only $t$.

Answer 2

To even write the ratios $\frac{\mathrm{d}v}{\mathrm{d}t}$, $\frac{\mathrm{d}v}{\mathrm{d}x}$, and $\frac{\mathrm{d}x}{\mathrm{d}t}$, you have to presuppose that each pair of variables is related (by a differentiable function).

So if it even makes sense to write those, then it's automatically true that $x$ and $\frac{\mathrm{d}v}{\mathrm{d}t}$ have a functional relationship, and similarly for the other two derivatives.

On a domain where we can additionally express the relationships as $v = f(x)$ and $t = g(x)$ for some differentiable functions $f$ and $g$, then we have the relations

$$ \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{f'(x)}{g'(x)} $$ $$ \frac{\mathrm{d}v}{\mathrm{d}x} = f'(x) $$ $$ \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{g'(x)} $$