Sine Inverse Trig Derivative

[Jason76]

Given:

\(\displaystyle \dfrac{dy}{dx}\arcsin(u) = \dfrac{1}{\sqrt{1 - u^{2}}}(\dfrac{du}{dx})\)

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

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

\(\displaystyle du = 3 - 12x^{2}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - (3x - 4x^{3})^{2}}}(3x - 12x^{2})\)

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

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1)}}\) I don't understand how the denominator in this line came about.

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2}) }{\sqrt{4x^{2} - 1 )^{2}(1 - x^{2}) }}\) I also don't understand the denominator in this line.

\(\displaystyle \dfrac{dy}{dx} = \pm \dfrac{3}{\sqrt{1 - x^{2}}}\) - Answer The denominator and numerator is a mystery.

 

[Deleted member 4993]

Given:

\(\displaystyle \dfrac{dy}{dx}\arcsin(u) = \dfrac{1}{\sqrt{1 - u^{2}}}(\dfrac{du}{dx})\)

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

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

\(\displaystyle du = 3 - 12x^{2}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - (3x - 4x^{3})^{2}}}(3x - 12x^{2})\)

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

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1)}}\) I don't understand how the denominator in this line came about.

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2}) }{\sqrt{4x^{2} - 1 )^{2}(1 - x^{2}) }}\) I also don't understand the denominator in this line.

\(\displaystyle \dfrac{dy}{dx} = \pm \dfrac{3}{\sqrt{1 - x^{2}}}\) - Answer The denominator and numerator is a mystery.

I don't understand your first line - however, the last three lines where you have problems, stem from algebra.

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2})}{\sqrt{(1 - x^{2})(16x^{4} - 8x^{2} + 1)}}\) I don't understand how the denominator in this line came about.

\(\displaystyle 1 - 9x^{2} + 24x^{4} - 16x^{6} \ = \ (1 - x^{2})(16x^{4} - 8x^{2} + 1)\)

The above is achieved through factorization and application of rational root theorem (http://www.purplemath.com/modules/rtnlroot.htm) - which would become obvious if you substitute k = x²

These are taught in algebra II - high school.

Next question:

\(\displaystyle \dfrac{dy}{dx} = \dfrac{3(1 - 4x^{2}) }{\sqrt{(4x^{2} - 1 )^{2}(1 - x^{2}) }}\) I also don't understand the denominator in this line.

this comes from the fact:

\(\displaystyle (16x^{4} - 8x^{2} + 1) \ = \ (4x^{2} - 1 )^{2} \) which comes from \(\displaystyle a^2 + b^2 + 2ab = (a+b)^2\)

Again - these are taught in algebra II - high school.

next question:

\(\displaystyle \dfrac{dy}{dx} = \pm \dfrac{3}{\sqrt{1 - x^{2}}}\) - Answer The denominator and numerator is a mystery

Eleminate common factors from the numerator and denominator - in this case (4x² - 1). - this is taught in Algebra I (middle school)

Several volunteers in this forum had advised you to shore up your knowledge of algebra before you tackle calculus. Please heed their advice. Otherwise you would be swimming upstream all the way and the journey will not be pleasant.[edit]