Why is $\frac{d}{dt}[\frac{1}{y}] = \frac{d}{dt}[\cos{x}]$ result in an application of the chain rule in related functions?

Question

Why does the below step result in a chain rule? Intuitively, I would just differentiate it against y and x resulting in $-y^{-2} = -\sin{x}$.

Khan Acad mentioned something about relating y = y(t) and x = x(t), but my brain is struggling to comprehend why it can just be related like this... (Though I could just be getting it completely wrong)

Answer 1

I actually think it is easier if you separate out differentiation from derivatives. Differentiation is the process of converting an equation of variables to an equation of differentials. A derivative is a ratio of derivatives. I think that separating these processes makes the whole of calculus easier to understand.

So, in your case, you have:

$$y^{-1} = \cos(x)$$

If we just differentiate it, we get:

$$ -y^{-2}\,dy = -\sin(x)\, dx$$

Here, no variable has a privileged place. We aren't differentiating with respect to anything. Every expression is followed by the differential of the variables.

The question wants $\frac{dy}{dt}$. To do that, start by solving for $dy$:

$$ dy = \sin(x)\,y^2\,dx $$

Now, just divide both sides by $dt$ (you can do anything you want to an equation, as long as you do it to both sides - even dividing by a brand new differential):

$$\frac{dy}{dt} = \sin(x)\,y^2\,\frac{dx}{dt}$$

Answer 2

It might be somewhat complicated, but the essence of the problem is simple.

We know that the variable $x, y$ is unknown functions of $t$, In other words, we know that the some functions $f(t), g(t)$ exist such that $x = f(t), y = g(t)$ but we don't know $yet$ what the actual expression of $f, g$ look like. It maybe look like $f(t) = 2t$, or $f(t) = \tan(t) + 12$ or anything can be. But thanks to the problem, we know some relation between $f$ and $g$.

We know $\frac{1}{y} = \cos(x)$, can be written as $\frac{1}{g(t)} = \cos(f(t))$. You can see that there is no $x, y$ in the relation. So you can just simply differentiate it with $t$(because LHS and RHS are always equal).

According to your idea, just simply differentiate LHS and RHS with $x$, can be also applied, but we can't get $\frac{dy}{dt}$, which we want. Because: $$\frac{1}{y} = \cos{x}$$ If we differnetiate both side with $\frac{d}{dx}$: $$-\frac{1}{y^2}\frac{dy}{dx}=-\sin(x)\frac{dx}{dx}$$ Which is same as: $$\frac{dy}{dx}=y^2 \sin(x)$$

Answer 3

I don't actually think that it is the chain rule. Since we have the value for $\dfrac 1y$ as $\cos{x}$, you can directly substitute it, if you are looking for $\dfrac{d}{dt}[\dfrac 1y]$. (I say this about the title, to make it clearer)

Here's another solution I tried to make. Hope you find this one helpful.

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$\because \dfrac{dx}{dt} = -2, dx = -2dt \implies \int{dx} = \int{-2dt} =\implies x = -2t + C$ (preparing stuff for application in chain rule)

(If the correct answer is $0$, let's take $C = 0$, which makes $x = -2t$. Also, it (taking $C = 0$) might work, since no special graph to find the intercept has been given or because of lack of reference to anything that alters the value of $C$. )

$\implies \cos{x} = \cos{(-2t)}$

$\therefore , \dfrac 1y = \cos{x} = \cos(-2t), y = \dfrac 1{\cos{(-2t)}}$

$\therefore, \dfrac{dy}{dt} = \dfrac{d(\dfrac 1{\cos{(-2t)}})}{dt} = \dfrac{-\sin{(-2t)}}{\cos^2{(-2t)}}\times-2 = \dfrac{2\sin(-2t)}{\cos^2{(-2t)}}$ (Here's where we use the chain rule, alongside the quotient rule.)

Now we are looking for the derivative at $t = -\dfrac \pi2$ , that is , $x = \pi$ (since $x = -2t$ as we get from the integration above), so just plug in the values, you get $\dfrac {dy}{dt} = 0$.

For an understandable solution, since we have $x = -2t$, we can write $\dfrac{dy}{dt} = \dfrac{2\sin(-2t)}{\cos^2{(-2t)}}$ as $-2\dfrac{dy}{dx} = \dfrac{2\sin(x)}{\cos^2(x)}$ (we have to remove the $-2$ factor in $x$). Now plug in $x = \pi$ and you get $\dfrac{dy}{dx} = 0$ $\implies \dfrac{dy}{dt} = -2\dfrac{dy}{dx} = 0$