Derivatives of Inverse Trig Functions #2

[Silvanoshei]

\(\displaystyle f(x)=sin^{-1}\sqrt{(1-x^{2}})\)

My Work:

\(\displaystyle f(x)=\frac{1}{\sqrt{1-u^{2}}}*\frac{d}{dx}(1-x^{2})^{\frac{1}{2}}\)

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-((1-x^{2})^\frac{1}{2})^{2}}}*\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}(-2x)?\)

Is it ok to leave it in that form? Reducing it anymore seems like a headache lol.

 

[Deleted member 4993]

\(\displaystyle f(x)=sin^{-1}\sqrt{(1-x^{2}})\)

My Work:

\(\displaystyle f(x)=\frac{1}{\sqrt{1-u^{2}}}*\frac{d}{dx}(1-x^{2})^{\frac{1}{2}}\)

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-((1-x^{2})^\frac{1}{2})^{2}}}*\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}(-2x)?\)

Is it ok to leave it in that form? Reducing it anymore seems like a headache lol.

It is okay - but the denominator reduces to 'x' and then cancels out with 'x' from '2x'.

 

[srmichael]

\(\displaystyle f(x)=sin^{-1}\sqrt{(1-x^{2}})\)

My Work:

\(\displaystyle f(x)=\frac{1}{\sqrt{1-u^{2}}}*\frac{d}{dx}(1-x^{2})^{\frac{1}{2}}\)

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-((1-x^{2})^\frac{1}{2})^{2}}}*\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}(-2x)?\)

Is it ok to leave it in that form? Reducing it anymore seems like a headache lol.

Eh, you may get some points off if you don't reduce this one as it is designed to have some terms cancel.

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-(1-x^{2})}}*\frac{-x}{\sqrt{1-x^{2}}}\)

\(\displaystyle f'(x)=\frac{1}{x}*\frac{-x}{\sqrt{1-x^{2}}}\)

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

 

[Deleted member 4993]

\(\displaystyle f(x)=sin^{-1}\sqrt{(1-x^{2}})\)

My Work:

\(\displaystyle f(x)=\frac{1}{\sqrt{1-u^{2}}}*\frac{d}{dx}(1-x^{2})^{\frac{1}{2}}\)

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-((1-x^{2})^\frac{1}{2})^{2}}}*\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}(-2x)?\)

Is it ok to leave it in that form? Reducing it anymore seems like a headache lol.

Different method:

\(\displaystyle f(x)=sin^{-1}\sqrt{(1-x^{2}})\)

sin(y) = √(1 - x²) → sin²(y) = 1 - x² and cos²(y) = x² → cos(y) = ±x

2* sin(y) * cos(y) * dy/dx = - 2x

dy/dx = ± 1/sin(y) = ±(1-x²) ^-1/2

 

[Silvanoshei]

Hmmm, so...

\(\displaystyle f'(x)=\frac{1}{(1-((1-x^{2})^\frac{1}{2})^{2})^\frac{1}{2}}*\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}(-2x)\)

turns into...

\(\displaystyle f'(x)=\frac{1}{\sqrt{1-(1-x^{2}})}*\frac{\frac{1}{2}}{\sqrt{(1-x^{2})}}*\frac{-2x}{1}\)

which is...

\(\displaystyle f'(x)=\frac{1}{x}*\frac{-x}{\sqrt{1-x^{2}}}\)

ahhhh...

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

That looks much better, thanks for the run-down!! :mrgreen: