Chain rule in relative rate of change?

Question

If I find the relative rate of change for

$y=x^4$

when $x,y$ are differentiable functions of $t$, I will get

$\frac{\mathrm{d} y }{\mathrm{d} t} = \frac{\mathrm{d} (x^4) }{\mathrm{d} t} =4x^3 \frac{\mathrm{d} x }{\mathrm{d} t}$

I'm confused about the $\frac{\mathrm{d} x }{\mathrm{d} t}$ part.

If, for example, $sin(x)$ is a function of $x$ then finding $\frac{\mathrm{d} (sin(x)) }{\mathrm{d} x}$ equals to $cos(x)\frac{\mathrm{d} x }{\mathrm{d} x}$ which equals to $cos(x)$

And in $y=x^4$ if $x$ is a function of $t$, then why doesn't

$\frac{\mathrm{d} x^4 }{\mathrm{d} t} = 4x^3 \frac{\mathrm{d} t }{\mathrm{d} t} $?

Answer 1

From the chain rule, we have: $$\frac {du}{dt}=\frac {du}{dv} \cdot \frac {dv}{dt}$$ Hence, $$\frac {dy}{dt}=\frac {d(x^4)}{dt}=\frac {d(x^4)}{dx} \cdot \frac {dx}{dt}=4x^3 \frac {dx}{dt}$$

In the case of $\frac {d(\sin x)}{dx}=\cos x$, chain rule need not be applied, and even if it is, will yield the same result since $\frac {dx}{dx}=1$.

Answer 2

If, for example, $sin(x)$ is a function of $x$ then finding $\frac{\mathrm{d} (sin(x))}{\mathrm{d} x}$ equals to $cos(x)\frac{\mathrm{d} x }{\mathrm{d} x}$ which equals to $cos(x)$

Sure, but that is differentiating with respect to $x$; instead consider that when $x$ is a function of $t$, then differentiating with respect to $t$ gives us:$$\begin{align}\dfrac{\mathrm d\sin(x) }{\mathrm d t}&=\dfrac{\mathrm d \sin(x)}{\mathrm d x}\dfrac{\mathrm d x}{\mathrm d t}\\[1ex]&= \cos(x)\cdot\dfrac{\mathrm d x}{\mathrm d t}\end{align}$$

Answer 3

For example, let $y=x^4, x=t^2$. Then: $$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}=\\ 4x^3\cdot 2t=4(t^2)^3\cdot 2t=8t^7$$ To verify: $$y=x^4=(t^2)^4=t^8 \Rightarrow \frac{dy}{dt}=8t^7.$$