# derivative of an inverse trig function

by efekwulsemmay
Tags: derivative, function, inverse, trig
 P: 53 1. The problem statement, all variables and given/known data $$y=sec^{-1}\frac{1}{t}, 0  P: 418 The step, [tex]\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}} = \frac{-1}{t\cdot\sqrt{1-t^{2}}}$$ is incorrect. Remember $$a\sqrt{b} = \sqrt{a^2b}$$ so you have, $$t\sqrt{1/t^2-1} = \sqrt{1-t^2}$$
 P: 53 ooooooooh ok. that makes more sense. thats clever hehehe thank you so much
HW Helper
P: 3,353

## derivative of an inverse trig function

For next time, you may remember that since $$\cos x = \frac{1}{\sec x}$$, it is true that $$\sec^{-1} \left(\frac{1}{x}\right) = \cos^{-1} x$$ and so the derivative is standard.

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