Derivatives with cot and tan

[javaking]

I had a take home test over break, and my math teacher put a problem on it thas was :

Find the derivative of y= cot-1 1/x - tan-1 x


I believe the problem is meant to be read y= (cot-1 1/x) - (tan-1 x)


I found the answer to be y= x- 1/1+x. I was wondering if this was the correct answer. Thanks.

 

[galactus]

I assume you mean arccot and arctan?.

The derivative of arctan(x) is \(\displaystyle \L\\\frac{1}{1+x^{2}}\)

The derivative of arccot(x) is \(\displaystyle \L\\\frac{-1}{1+x^{2}}\)

Therefore the derivative of arccot(1/x) is:

\(\displaystyle \L\\\frac{-1}{1+(\frac{1}{x})^{2}}(\frac{-1}{x^{2}})=\frac{1}{1+x^{2}}\)

Looks like you'll get a very simple answer.

 

[soroban]

Looks like you'll get a very simple answer.

Absolutely right, galactus!

Note that: \(\displaystyle \,\cot^{-1}\left(\frac{1}{x}\right)\;=\;\tan^{-1}(x)\)