How do I find the derivative of y=sin^-1 (3/(t^2))y=sin1(3t2)?

1 Answer
Apr 29, 2018

-6/(t^3sqrt(1-(9/t^4))6t31(9t4)

Explanation:

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))