# How do I find the derivative of y=sin^-1 (3/(t^2))?

Apr 29, 2018

-6/(t^3sqrt(1-(9/t^4))

#### Explanation:

Use the chain rule :

$\textcolor{p u r p \le}{f \left(g \left(t\right)\right) = {f}^{'} \left(g \left(t\right)\right) \cdot {g}^{'} \left(t\right)}$

$f = {\sin}^{-} 1$ here
$g = \frac{3}{t} ^ 2$ here

$\arcsin \left({\sin}^{-} 1\right) = \frac{1}{\sqrt{1 - {\left(\frac{3}{t} ^ 2\right)}^{2}}} \cdot \frac{d}{\mathrm{dt}} \left[\frac{3}{t} ^ 2\right]$

For $\frac{d}{\mathrm{dt}} \left[\frac{3}{t} ^ 2\right]$:

Pull out the 3 because it is a constant so

$3 \frac{d}{\mathrm{dt}} \left[\frac{1}{t} ^ 2\right]$

$\frac{1}{t} ^ 2$ is the same as saying ${t}^{-} 2$ of which it is easy to take the derivative using the power rule :

$\frac{d}{\mathrm{dt}} {t}^{-} 2 = - 2 {t}^{-} 3$

so we have $3 \cdot \left(- 2 {t}^{-} 3\right)$

Now you're almost done, simplify:

$= \frac{- 6 {t}^{-} 3}{\sqrt{1 - {\left(\frac{3}{t} ^ 2\right)}^{2}}}$

=-6/(t^3sqrt(1-(3/t^2)^2)

=-6/(t^3sqrt(1-(9/t^4))