The chain rule and squared trig functions

Question

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.

Answer 1

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

Answer 2

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 $2w=2\cot x=2\cot(\sin\theta)$.

Answer 3

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

Answer 4

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.