# Derivative of $h(t)= \sin (\cos^{-1}t$)?

Can someone please explain the steps/rules I need to preform to find the derivative of

$h(t)= \sin (\cos^{-1}t)$?

I tried to used the product rule, and realized it was obviously a failure. Using the chain rule I get:

$\frac{dy}{dx}= \cos(\cos^{-1}t)\frac{-t}{\sqrt{1-t^2}}$

I know I am not finished, what am I supposed to do with this?

• First, that should be dh/dt, not dy/dx. Secondly, why are you not finished? That's the derivative. – foobar1209 Mar 17 '14 at 4:37
• Well, the only thing you can apply here is the chain rule, of course: you have composition of two functions, not a product. You're indeed almost finished, just change that $\;-t\;$ to $\;-1\;$ and you're done. – DonAntonio Mar 17 '14 at 4:39
• Oops, I missed the mistake that @DonAntonio pointed out. It does need to be changed. – foobar1209 Mar 17 '14 at 4:41
• Okay, I see my mistake with the t needing to be 1, but I don't understand the rest of it because before coming here, I checked solutions to my book which showed: h(t)= sin (arccos t)= sqrt(1-t^2) therefore, h'(t)= [(1/2)(1-t^2)^(-1/2)][-2t]= -t/sqrt(1-t^2) – Mac Mar 17 '14 at 4:46
• Yes, the $\;-t\;$ now appears after $\;\cos\arccos t=t\;$ , not before... – DonAntonio Mar 17 '14 at 4:56

\begin{align} h'(t) &= \frac{d}{dt}\sin(\arccos t) \\ &=\cos(\arccos t) \frac{d}{dt}(\arccos t) & \text{chain rule}\\ &=\cos(\arccos t) \cdot -\frac{1}{\sqrt{1-t^2}} & \text{$\frac{d}{dt}(\arccos t)=-\frac{1}{\sqrt{1-t^2}}$}\\ &=t \cdot -\frac{1}{\sqrt{1-t^2}} & \text{$\cos(\arccos t)=t$}\\ &= \frac{-t}{\sqrt{1-t^2}} \end{align}

You can also make the observation that $h(t)=sin(cos^{-1}(t)) = \sqrt{1-t^2}$ and differentiate from there (do some trig to see where this observation comes from).

\begin{align} (\sqrt{1-t^2})' = \frac{1}{2\sqrt{1-t^2}}\times(1-t^2)'=\frac{-2t}{2\sqrt{1-t^2}}=\frac{-t}{\sqrt{1-t^2}} \end{align}

Use the chain rule:

\begin{align}h'(t) &= \cos(\cos^{-1}t)\cdot\frac{d}{dt}(\cos^{-1}t)\\ &=t\cdot\frac{-1}{\sqrt{1-t^2}}\\ &=\frac{-t}{\sqrt{1-t^2}}\end{align}

HINT:

If $\displaystyle\cos^{-1}t=u,\cos u =t$

As the Principal values of inverse cosine ratio lies in $\displaystyle\left[0,\pi\right]$

$\displaystyle\sin(\cos^{-1}t)=\sin u\ge0\implies \sin u=+\sqrt{1-\cos^2u}=\sqrt{1-t^2}$

• This is exactly what I wasn't understanding from my book as it doesn't explain what changes are being made. I didn't make this connection, Thanks :) – Mac Mar 17 '14 at 5:06
• For what it's worth, compositions of the form $\operatorname{trig}_1 \left( \operatorname{trig}_2^{-1} (t) \right)$, where $\operatorname{trig}_1$ and $\operatorname{trig}_2$ are any of the six trigonometric functions, are always equivalent to rational expressions, whose numerators and denominators are expressions of the form $1$, $t$, $\sqrt{1 + t^2}$, or $\sqrt{1 - t^2}$. I call these Pythagorean expressions, as they form the labels of the sides of a right triangle. – Sammy Black Mar 17 '14 at 5:46
• The previous comment is for the OP. – Sammy Black Mar 17 '14 at 5:47

If you don't like the function composition, we can also re-arrange this as $\ \arcsin (h) \ = \ \arccos (t) \ \$ and use implicit differentiation (along with the known derivatives of these inverse trigonometric functions) to write

$$\frac{d}{dt} \ [\arcsin (h)] \ = \ \frac{d}{dt} \ [\arccos (t)]$$

$$\Rightarrow \ \ \frac{1}{\sqrt{1 \ - \ h^2}} \ \cdot \ \frac{dh}{dt} \ = \ -\frac{1}{\sqrt{1 \ - \ t^2}} \ \ \Rightarrow \ \ \frac{dh}{dt} \ = \ -\frac{\sqrt{1 \ - \ h^2}}{\sqrt{1 \ - \ t^2}} \ \ .$$

But since $\ h(t) \ = \ \sin (\arccos (t)) \ ,$ we can apply the Pythagorean Identity to write (with appropriate domain restrictions)

$$\frac{dh}{dt} \ = \ -\frac{\cos (\arccos (t))}{\sqrt{1 \ - \ t^2}} \ = \ -\frac{t}{\sqrt{1 \ - \ t^2}} \ \ .$$