# The chain rule and squared trig functions

I am trying to do this homework problem and I have no idea how to approach it. I have tried many methods, all resulting in failure. I went to the books website and it offers no help. I am trying to find the derivative of the function $$y=\cot^2(\sin \theta)$$

I could be incorrect but a trig function squared would be the result of the trig function with the angle value and then squared. Not the angle value squared, that would give a different answer. Knowing this I also know that I can not use the table of simple trig derivatives so I know I can't just take the derivative as $$y=\cot^2(x)$$ $$x=\sin(\theta)$$

This does not help because I can't get the derivative of cot squared. What I did try to do was rewrite it as $\frac{\cos x}{\sin x}\frac{\cos x}{\sin x}$ and then find the derivative of that but something went wrong with that and it does not produce an answer that is like the one in the book. In fact the book gets a csc squared in the answer so I know they are doing something very different.

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Use the power rule. – Américo Tavares Feb 15 '12 at 20:15
$$y = cot^2(sin\theta)$$ $$y' = 2.cot(sin\theta).(-csc^2(sin\theta)).cos(\theta)$$ – Inquest Feb 15 '12 at 20:16
What is a power rule? – Jordan Feb 15 '12 at 20:17
It is the differentialion rule $(u^{n})^{\prime }=nu^{n-1}u^{\prime }$. – Américo Tavares Feb 15 '12 at 20:21
... and then use the chain rule. – Américo Tavares Feb 15 '12 at 20:27

## 4 Answers

Indeed, $\cot^2(a)$ means $$\left(\cot (a)\right)^2.$$

You need to apply the Chain Rule twice: first, to deal with the square: set $g(u)=u^2$ as your "outside function", and $u=f(\theta) = \cot(\sin(\theta))$ as your inside function. Since $g'(u) = 2u$, then $$\frac{d}{d\theta}\cot^2(\sin(\theta)) = \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2 = g'(u)f'(\theta) = 2uf'(\theta) = 2\cot\bigl(\sin(\theta)\bigr)f'(\theta).$$ Now let's deal with $f'(\theta)$; we have $f(\theta) = \cot\bigl(\sin(\theta)\bigr)$. The "outside function" is $h(u) = \cot(u)$, the "inside function" is $u(\theta) = \sin(\theta)$. Since $h'(u) = -\csc^2(u)= -\left(\csc(u)\right)^2$, and $f'(\theta) = \cos(\theta)$, we have: $$\frac{d}{d\theta}\cot\bigl(\sin(\theta)\bigr) = h'(u)u'(\theta) = -\csc^2(\sin\theta)\cos(\theta).$$

Putting it all together: \begin{align*} \frac{d}{d\theta}\cot^2\bigl(\sin(\theta)\bigr) &= \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \left(-\csc^2\left(\sin(\theta)\right)\left(\frac{d}{d\theta}\sin(\theta)\right)\right)\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot\left(-\csc^2\left(\sin(\theta)\right)\cos(\theta)\right)\\ &= -2\cot\bigl(\sin(\theta)\bigr)\csc^2\bigl(\sin(\theta)\bigr)\cos(\theta). \end{align*}

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So technically to find the derivative of $sin3x$ I have to use the chain rule? I don't understand why I have to use the chain rule twice I guess. – Jordan Feb 15 '12 at 20:26
@Jordan: $\left(\cot(\sin(\theta))\right)^2$ is a composition of three functions: first you compute $\sin(\theta)$; then you take the output of that and "feed it into" $\cot$ to get $\cot(\sin(\theta))$. Then you take the ouput of that and feed it into the square, to get $\left(\cot(\sin(\theta))\right)^2$. In total, you've done two compositions, (you've twice taken the output of one function and used it as the input for another function). Each composition gives you one application of the Chain Rule when doing the derivative. – Arturo Magidin Feb 15 '12 at 20:28
@Jordan: Yes, $(\sin(3x))'$ is computed using the Chain Rule: $(\sin(3x))' = \cos(3x)\cdot(3x)' = 3\cos(3x)$. – Arturo Magidin Feb 15 '12 at 20:29

You started in a way that leads to the answer.

Let $y=\cot^2(\sin \theta)$. We want to find $\dfrac{dy}{d\theta}$. Make the substitution $x=\sin\theta$. (Comment: when we are using substitution, it is more common to use letters like $u$, $v$, $w$, but $x$ is fine here.) Note that $$y=\cot^2 x.$$ Differentiate with respect to $\theta$. We get $$\frac{dy}{d\theta}=\frac{dx}{d\theta}\frac{dy}{dx}.$$

Easily, we have $\dfrac{dx}{d\theta}=\cos\theta$. We still need to find $\dfrac{dy}{dx}$ where $y=\cot^2 x$.

How shall we do this? There are several possible ways. For example, $\cot^2 x$ is a product, so we could use the Product Rule. Or else, we can use the Chain Rule again. Let $w=\cot x$. Then $$y=w^2,$$ and therefore $$\frac{dy}{dx}=\frac{dw}{dx}\frac{dy}{dw}.$$ Since $y=w^2$, we have $\dfrac{dy}{dw}=2w$. We still need $\dfrac{dw}{dx}$.

Since $w=\cot x$, in order to find $\dfrac{dw}{dx}$ we need to find the derivative of $\cot x$. There are many approaches to this. Perhaps it is one of the derivatives that you just remember: the answer is $-\csc^2 x$. Or if you don't remember this derivative, use the fact that $$\cot x=\frac{\cos x}{\sin x}$$ and use the Quotient Rule. After a while, we find that $$\frac{dw}{dx}=\frac{-(\sin^2 x+\cos^2 x)}{\sin^2 x}.$$ Using the fact that $\sin^2 x+\cos^2 x=1$, you can simplify this to $-\dfrac{1}{\sin^2 x}$, and then to $-\csc^2 x$. Finally, it is time to put the pieces together. We had $$\frac{dy}{d\theta}=\frac{dx}{d\theta}\frac{dy}{dx}=\frac{dx}{d\theta}\frac{dw}{dx}\frac{dy}{dw}.$$

In the calculations above, we found: $$\frac{dx}{d\theta}=\cos\theta;\qquad \frac{dw}{dx}=-\csc^2 x;\qquad \frac{dy}{dw}=2w.$$ Multiply them together, but first express everything in terms of the original variable $\theta$. So $-\csc^2 x=-\csc^2(\sin\theta)$ and $2w=2\cot x=2\cot(\sin\theta)$.

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@Arturo Magidin: Thank you. Of course, what I typed was precisely your solution, apart from the fact that yours did not have the typo. The reason I did it is that the OP may find the substitution language more familiar. – André Nicolas Feb 15 '12 at 21:53
Out of curiosity: is it standard where you are (France?) to write the substitution rule as $\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}$, as you do in the post? My impression is that in the US it is almost invariably written $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$. I also ask because it gives a different "mental picture", I think. You are working from the inside out, whereas I tend to tell my students that when you apply the Chain Rule you work "outside in" (or LIFO: last function in [in evauating], first function out [in computing derivatives]). – Arturo Magidin Feb 15 '12 at 21:59
@Arturo Magidin: It is Canada. And the order is an idiosyncrasy. The main reason I adopted that pattern is that at the very beginning level, the standard rookie error is to forget about taking the derivative of the "inside" function. – André Nicolas Feb 15 '12 at 22:18
Curious. For me, the standard "rookie error" is to do both derivatives at the same time. E.g., $\frac{d}{dx}(\sin x)^2 \to 2\cos x$. – Arturo Magidin Feb 16 '12 at 3:59

$y'=2(\cot(\sin \theta))(\cot(\sin \theta))'\cdot (\sin \theta)'=2 \cdot (\cot(\sin \theta))(-1-\cot^2(\sin \theta))\cdot \cos \theta$

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This isn't really helpful for me to see at all. My book also has the answer to this problem. – Jordan Feb 15 '12 at 20:18

The way I was taught allows me to use the chain rule without really needing to think about it. This would be the way I'd write it:

$dy=d(\cot^2(\sin\theta))=2\cot(\sin\theta)d(\cot(\sin\theta))=2\cot(\sin\theta)(-\csc^2(\sin\theta))d(\sin\theta)=$

$-2\cot(\sin\theta)\csc^2(\sin\theta)\cos\theta d\theta$

Now what did I do to get from one step to the next? Everything inside the d() is a substitution. So my first substitution was

$\frac{d\cot^2(\sin\theta)}{d\theta}=\frac{du^2}{du}\frac{du}{d\theta}=2u\frac{du}{d\theta}=2\cot(\sin\theta)\frac{d\cot(\sin\theta)}{d\theta}$

After a while, the substitutions come without thinking. First I see I need the derivative of the square of some function, then the derivative of the cotangent of some function, etc.

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