Question

Totally lost on this one..
derivative of y=sec^-1(1/9t^4) 0 13 years ago

Answer 1

\[\frac{d(sec^{-1}(x))}{dx}=\frac{1}{x\sqrt{x^2-1}}\]

Answer 2

Or for your purposes:
\[\frac{d(sec^{-1}(x))}{dt}=\frac{\frac{dx}{dt}}{x\sqrt{x^2-1}};\ \ x=\frac{1}{9t^4};\ \ \frac{dx}{dt}=-\frac{4}{9t^5}\]

Answer 3

I know the formula I just get stuck with arithmetic. Was looking at wolfram, it brings the x into the radical..not sure how to that