Inverse Trig Derivative Problem about ArcSin

**Jason76** · November 8th 2012, 08:51 AM

Find the Derivative

$y = 2 \arcsin \sqrt{1 - 2x}$

using the formula for derivative of arcsin:

$\frac{d}{dx}\sin^{-1}$

and

$u = (\frac{1}{\sqrt{1 - u^{2}}})(\frac{du}{dx})$

Step 1.

$y = 2[\frac{1}{\sqrt{1-(\sqrt{1-2x})^{2}}}] [\frac{1}{2}(1-2x)^{-1/2}](-2)$

Step 2:

Can't understand how they get from here (below this sentence) to the answer.

$y = (\frac{1}{\sqrt{1-1+2x}})(\frac{1}{\sqrt{1-2x}})(-2)$

Answer:

$y = \frac{-2}{\sqrt{2x - 4x^{2}}}$

**fkf** · November 8th 2012, 09:02 AM

Multiplying sqrt(a)*sqrt(b) = sqrt(a*b)

In this case we have the denominator

sqrt(1-1+2x)*sqrt(1-2x) = sqrt(2x)*sqrt(1-2x) = sqrt((2x)*(1-2x)) = sqrt(2x-4x^2)

**topsquark** · November 8th 2012, 09:05 AM

Originally Posted by Jason76

Can't understand how they get from here (below this sentence) to the answer.

$y = (\frac{1}{\sqrt{1-1+2x}})(\frac{1}{\sqrt{1-2x}})(-2)$

Answer:

$y = \frac{-2}{\sqrt{2x - 4x^{2}}}$

The notation in equation 3 was a little confusing, but no matter. It's clear what you meant.

Note:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a}{b}$

Here we have a = 2x and b = 1 - 2x. Can you finish now?

-Dan

Ya beat me fkf!

**Jason76** · November 8th 2012, 02:41 PM

Originally Posted by topsquark

The notation in equation 3 was a little confusing, but no matter. It's clear what you meant.

Note:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a}{b}$

Here we have a = 2x and b = 1 - 2x. Can you finish now?

-Dan

Ya beat me fkf!

Yes, thanks a lot. But I'm wondering, this was an example of one number inside of a radical being multiplied by two numbers in another radical. However, what if there were two numbers in one radical being multiplied by two in another (and similar patterns)?

**topsquark** · November 8th 2012, 03:52 PM

Originally Posted by Jason76

Yes, thanks a lot. But I'm wondering, this was an example of one number inside of a radical being multiplied by two numbers in another radical. However, what if there were two numbers in one radical being multiplied by two in another (and similar patterns)?

If I am reading this correctly you are looking for $\sqrt{cd} \cdot \sqrt{ef} = \sqrt{cdef}$ .

It's similar to the equation I posted earlier. Just let a = cd and b = ed.

If I am wrong about what you are asking, please let us know.

-Dan

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