Derivatives of Inverse Trig Functions

[Silvanoshei]

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

My work:

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

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

\(\displaystyle f'(x)=(\frac{-1}{1+4x^{-2}})*(\frac{-2}{x^{2}})\)

\(\displaystyle f'(x)=(\frac{2}{x^{2}+4x})?\)

 

[srmichael]

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

My work:

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

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

\(\displaystyle f'(x)=(\frac{-1}{1+4x^{-2}})*(\frac{-2}{x^{2}})\)

\(\displaystyle f'(x)=(\frac{2}{x^{2}+4x})?\)

Alllllllmost.

\(\displaystyle f'(x)=\frac{2}{x^{2}+4}\)

Btw, thanks for clearly showing your work!

 

[Silvanoshei]

Alllllllmost.

\(\displaystyle f'(x)=\frac{2}{x^{2}+4}\)

Btw, thanks for clearly showing your work!

Whoops! Yeah, totally forgot those would cancel lol, was just thinking about the exponents. Thanks!