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

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Answer 1 · 2018-04-29T05:17:47

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

Use the **chain rule ** :

$color(purple)(f(g(t))=f^'(g(t))*g^'(t))$

$f=sin^-1$ here
$g= 3/t^2$ here

$arcsin (sin^-1)= 1/(sqrt(1-(3/t^2)^2))*d/dt[3/t^2]$

For $d/dt[3/t^2]$ :

Pull out the 3 because it is a constant so

$3 d/dt [1/t^2]$

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

$d/dt t^-2 = -2t^-3$

so we have $3*(-2t^-3)$

Now you're almost done, simplify:

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

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

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

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