Thread: Help with Derivative problem involving ARCSIN

1. Help with Derivative problem involving ARCSIN

Directions: Find the derivative of y with respect to the appropriate variabl.

y=sin^-1(2t)^1/2

Sin^-1 is the ARCSIN, and is the part I'm having trouble with. The answer is Square root of 2 over the square root of (1 - 2t^2), but I need help on how to obtain that answer. Thanks for your help!

2. Originally Posted by Alex225
Directions: Find the derivative of y with respect to the appropriate variabl.

y=sin^-1(2t)^1/2

Sin^-1 is the ARCSIN, and is the part I'm having trouble with. The answer is Square root of 2 over the square root of (1 - 2t^2), but I need help on how to obtain that answer. Thanks for your help!
The derivative of $\displaystyle y = \sin^{-1} (2t)^{1/2}$ (what you posted) is NOT $\displaystyle \frac{\sqrt{2}}{\sqrt{1 - 2t^2}}$.

For the given answer to be correct the question should be find the derivative of $\displaystyle y = \sin^{-1} (2^{1/2} t)$.

In which case you let $\displaystyle u = 2^{1/2} t$ and use the chain rule. You should know that if $\displaystyle y = \sin^{-1} u$ then $\displaystyle \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}$ ....