Chain rule mismatch

Question

Let $$ g\bigl(h(t)\bigr) = \cos(\sqrt{t}) $$ and $$ g(t) = \cos(t) $$ and $$ h(t) = \sqrt{t}. $$ Verify that $$ \frac{dg}{dh} = \frac{\;\frac{dg}{dt}\;}{\frac{dh}{dt}}. $$

I tried doing this but I got $$ \frac{dg}{dh} = -\sin(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}} $$ but that is not equal to $$ \frac{\frac{dg}{dt}}{\frac{dh}{dt}} = -2\sin(t) \cdot \sqrt{t} $$

Where did I make a mistake? Edit: I see the mistake but I tried another method and I don't know why that is wrong. $$\frac{dg}{dh}=-sin\left(h\right)\cdot \frac{dh}{dh}=-sin\left(\sqrt{t}\right)$$

Answer 1

The first derivative you found is $\dfrac{dg(h(t))}{dt}$, whereas you must compute $\dfrac{dg}{dh}$.

Since $h(t)=\sqrt t$, you have $g(t) = \cos(h(t)^2)$, and so

$$\frac{dg}{dh} = -\sin(h(t)^2) \cdot 2 h(t) = -2\sqrt t\,\sin(t)$$

On the other hand, by the chain rule,

$$\frac{dg}{dh} = \frac{\frac{dg}{dt}}{\frac{dh}{dt}} = \frac{-\sin(t)}{\frac1{2\sqrt t}} = -2\sqrt t\,\sin(t)$$

Answer 2

$\frac{dg}{dh}=\frac{d}{dh}g(t)$ where $h=\sqrt{t}$, $h^2=t$

so $\frac{dg}{dh}=\frac{d}{dh}g(h^2)=\frac{d}{dh}\cos(h^2)=-\sin(h^2)\cdot 2h=-\sin(t)\cdot 2\sqrt{t}$

Answer 3

Your functions are $g(t) = \cos(t)$ and $h(t) = \sqrt{t}$. Now you wish to find the derivative of $g$ with respect to $h$. So we can figure it out like this: $$ \begin{align} \frac{dg}{dh} &= \frac{d}{d(\sqrt{t})}(\cos(t))\\ &=\frac{d}{d(\sqrt{t})}[\cos((\sqrt{t})^2)]\\ \text{Let} \ \sqrt{t} = u\text{,} & \\[4pt] \therefore \frac{dg}{dh} &=\frac{d}{d(u)}[\cos(u^2)]\\ &= -2u\cdot \sin(u^2) \end{align} $$ Replacing $u = \sqrt{t}$, we get $$\frac{dg}{dh} = -2\sqrt{t} \cdot \sin(t)$$

Now, I believe you were able to find the RHS so I am not showing it here. Hope this helps.

Answer 4

To solve your problem here, you need to write $g$ in terms of $h$, i.e. if $$g(t)=\cos(t)$$ and: $$h(t)=\sqrt{t}$$ then: $$g(h)=\cos(h^2)$$ Then: $$\frac{dg}{dh}=-2\sin(h^2)h=-2\sin(t)\sqrt{2}$$ We see that: $$\frac{dg}{dt}=-\sin(t)$$ and: $$\frac{dh}{dt}=\frac{1}{2\sqrt{t}}$$ Hence: $$\frac{\frac{dg}{dt}}{\frac{dh}{dt}}=-2\sin(t)\sqrt{t}$$ as desired.

Edit:

So the main issue here is that you are mildly conflating notation, which is understandable in this case. If we first write that:

$$g(h)=\cos(h)$$

then it is absolutely true that:

$$\frac{dg}{dh}=-\sin(h)$$

Furthermore, if we have two function $g:\mathbb{R}\rightarrow \mathbb{R}$, and $h:\mathbb{R}\rightarrow \mathbb{R}$ (these are not the proper domains/ranges of the two specific functions, but you can easily modify this), then we can compose the two functions to obtain a new function $g\circ h:\mathbb{R}\rightarrow \mathbb{R}$. The chain rule then states that the derivative is given by:

$$\frac{d(g\circ h)}{dt}=\frac{dg}{dh}\frac{dh}{dt}$$

However, this is not what you are asking in your post. What you are starting with is two functions $g(t)=\cos(t)$ and $h(t)=\sqrt(t)$. If we fix these two functions, then we can not simply write: $$\frac{dg}{dh}=-\sin(h)$$ because $g$ is a function of $t$ not $h$. We can think of $dg/dh$ in this case as asking how does $g(t)$ change as $h(t)$ changes. To do this, we need to be able to take the derivative of $g(t)$ with respect to $h$, i.e. we need to rewrite the $t$ coordinate in terms of $h$. Once you do that, it all works out.

This is I think mildly easier to comprehend in the context of parametric curves. Suppose we have a map $f:\mathbb{R}\rightarrow \mathbb{R}^2$ where: $$f(t)=(h(t),g(t))$$ Here, we can think of the $h$ as the $x$ coordinate, and $g$ as the $y$ coordinate of the curve. If you graph this out you'll see you get the graph of a function, so what $dg/dh$ is finding, is more like the usual derivative $dy/dx$. But to calculate such a derivative, we need to write $y$ in terms of $x$, or else how are we going to differentiate some function of $t$ with respect to $x$?

Does that clear things up?