Another Factoring Inverse Trig Problem

[Jason76]

Finding the Derivative:

\(\displaystyle y = \arcsin(3x - 4x^{3})\)

Step 2)

\(\displaystyle y = \frac{1}{\sqrt{1 - (3x - 4x^{3})^{2}}} (3x - 4x^{3})\)

How did it get from step 2 to step 3. Plus don't understand the other steps below completely.

Step 3)

\(\displaystyle y = \frac{3-12x^{2}}{\sqrt{1- 9x^{2} + 24x^{4} - 16x^{6}}}\)

Step 4)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1} }\)

Step 5)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{4x^{2} - 1)^{2}(1 - x)^{2} }}\)

Step 6)

\(\displaystyle y = \pm\frac{3}{\sqrt{1 - x^{2}}}\) Final Answer

 

[mmm4444bot]

You need to learn about factoring polynomials and simplifying radicals, before attempting calculus exercises.

You continue to post questions on the calculus board about algebra. Why are you trying to learn calculus before you learn algebra?

 

[Deleted member 4993]

Finding the Derivative:

\(\displaystyle y = \arcsin(3x - 4x^{3})\)

Step 2)

\(\displaystyle y = \frac{1}{\sqrt{1 - (3x - 4x^{3})^{2}}} (3x - 4x^{3})\)

I do not understand - how did you go to step 2 from the given problem!!

Please explain the genesis of step (2)

I can show you how to go to step (3) from the given problem - but not through step (2)

How did it get from step 2 to step 3. Plus don't understand the other steps below completely.

Step 3)

\(\displaystyle y = \frac{3-12x^{2}}{\sqrt{1- 9x^{2} + 24x^{4} - 16x^{6}}}\)

Step 4)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1} }\)

Step 5)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{4x^{2} - 1)^{2}(1 - x)^{2} }}\)

Step 6)

\(\displaystyle y = \pm\frac{3}{\sqrt{1 - x^{2}}}\) Final Answer

.

 

[HallsofIvy]

Finding the Derivative:

\(\displaystyle y = \arcsin(3x - 4x^{3})\)

Step 2)

\(\displaystyle y = \frac{1}{\sqrt{1 - (3x - 4x^{3})^{2}}} (3x - 4x^{3})\)

This is incorrect. Yes, the derivative of arcsin(x) is \(\displaystyle \frac{1}{\sqrt{1- x^2}}\) so the derivative of \(\displaystyle arcsin(3x- 4x^3)\) is \(\displaystyle \frac{1}{\sqrt{1- (3x-4x^3)^2}}\) times the derivative of \(\displaystyle 3x- 4x^3\), not \(\displaystyle 3x- 4x^3\) itself. You should have
\(\displaystyle y'= \frac{1}{\sqrt{1- (3x- 4x^3)^2}}(3- 12x^2)\)

(A minor problem is that you are consistently writing "y= " when you mean "y'= ".)

How did it get from step 2 to step 3. Plus don't understand the other steps below completely.

Step 3)

\(\displaystyle y = \frac{3-12x^{2}}{\sqrt{1- 9x^{2} + 24x^{4} - 16x^{6}}}\)

Step 4)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1} }\)

Step 5)

\(\displaystyle y = \frac{3(1-4x^{2})}{\sqrt{4x^{2} - 1)^{2}(1 - x)^{2} }}\)

Step 6)

\(\displaystyle y = \pm\frac{3}{\sqrt{1 - x^{2}}}\) Final Answer