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?
On
Use the chain rule:
$$\begin{align}h'(t) &= \cos(\cos^{-1}t)\cdot\frac{d}{dt}(\cos^{-1}t)\\ &=t\cdot\frac{-1}{\sqrt{1-t^2}}\\ &=\frac{-t}{\sqrt{1-t^2}}\end{align}$$
On
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}
On
HINT:
If $\displaystyle\cos^{-1}t=u,\cos u =t$
As the Principal values of inverse cosine ratio lies in $\displaystyle\left[0,\pi\right]$
$\displaystyle\sin(\cos^{-1}t)=\sin u\ge0\implies \sin u=+\sqrt{1-\cos^2u}=\sqrt{1-t^2}$
On
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)
$$ \frac{dh}{dt} \ = \ -\frac{\cos (\arccos (t))}{\sqrt{1 \ - \ t^2}} \ = \ -\frac{t}{\sqrt{1 \ - \ t^2}} \ \ . $$
\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}