Help with Derivative problem involving ARCSIN

**Alex225** · October 30th 2008, 07:28 PM

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!

**mr fantastic** · October 31st 2008, 03:04 AM

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 $y = \sin^{-1} (2t)^{1/2}$ (what you posted) is NOT $\frac{\sqrt{2}}{\sqrt{1 - 2t^2}}$ .

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

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

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