Factoring Inverse Trig Problem

[Jason76]

Find the Derivative

\(\displaystyle y = \arccos(1-2x)\)

\(\displaystyle y = \frac{-1}{\sqrt{1 - (1-2x)^{2}}}(-2) \)

\(\displaystyle y = \frac{2}{\sqrt{1-1 + 4x - 4x^{2}}}\)

\(\displaystyle y = \frac{2}{\sqrt{4x-4x^{2}}}\)

Don't understand how it gets to the final answer, but understand everything above this sentence.

\(\displaystyle y = \pm\frac{1}{\sqrt{x-x^{2}}}\) Final Answer

 

[mmm4444bot]

Factor the radicand

4(x - x^2)

Next, simplify the radical. The square root of 4 is 2.

Now cancel that factor 2 in denominator with factor 2 in numerator.

 

[Jason76]

Factor the radicand

4(x - x^2)

Next, simplify the radical. The square root of 4 is 2.

Now cancel that factor 2 in denominator with factor 2 in numerator.

\(\displaystyle \frac{2}{\sqrt{4(x - x^{2})}}\)

\(\displaystyle \frac{2}{\sqrt{2(x - x^{2})}}\)

\(\displaystyle \frac{1}{\sqrt{1(x - x^{2})}}\)

\(\displaystyle \pm\frac{1}{\sqrt{x - x^{2}}}\)

But where did the \(\displaystyle \pm\) come from? Is it something to do with a parabola graph, considering the x squared?

Sorry about not putting La-Tex sign for derivative. Can't find it.

 

[mmm4444bot]

But where did the \(\displaystyle \pm\) come from?

Sorry about not putting La-Tex sign for derivative. Can't find it.

Who told you that the derivative could be negative? Did you consider the domain of y?

You may type an apostrophe for the prime symbol (y').

 

[Jason76]

Who told you that the derivative could be negative? Did you consider the domain of y?

You may type an apostrophe for the prime symbol (y').

The new book I got from the bookstore said this problem's answer was positive and negative.