is the derivative of arctan

[afcwestwarrior]

1/(x^2+1)

[nicksauce]

Yes.

[afcwestwarrior]

ok thanks

[NoMoreExams]

You can look up a list of derivatives and integrals of common functions on wikipedia or on google. But suppose you had to prove this by hand:

Let

y = arctan(x) \rightarrow tan(y) = x

Now use implicit differentiation to get

\frac{y'}{cos^{2}(y)} = 1 \rightarrow y' = cos^{2}(y) = cos^{2}(arctan(x))

Now draw your triangle to figure out what that expression becomes.

Note that you would do that by saying opp/adj = x so let opp = x and adj = 1 which means hyp = sqrt(1 + x^2), now since we need cosine = adj/hyp = 1/sqrt(1 + x^2) however since we have cosine^2, that gets rid of the sqrt( ) and you are left with 1/(1 + x^2). Hope this helped

[snipez90]

Or note that as NoMoreExams had y = arctan(x) \Rightarrow tan(y) = x. Now the derivative is sec^2(y)y' = 1 \Rightarrow y' = \frac{1}{sec^{2}(y)} = \frac{1}{tan^2(y)+1} = \frac{1}{x^2 + 1}.

[HallsofIvy]

1/(x^2+1)

Yes.
Strictly speaking it is 1/(x2+ 1)+ C

[Cvan]

Strictly speaking it is 1/(x2+ 1)+ C

Why do you need the constant for differentiation? Wouldn't that introduce the variable again when integrating back?

[nicksauce]

Strictly speaking it is 1/(x2+ 1)+ C

I've never heard of the constant of differentiation.

[transgalactic]

where there are no borders to the integral we are solving we use +C

but when we use a derivative i dont think we use +C

[rbj]

Strictly speaking it is 1/(x2+ 1)+ C

Halls, you're kidding, aren't you?

[afcwestwarrior]

thanks it turned out tan^-1=arctan

[mathwonk]

this equivalent to the fact that the derivative of tan is sec^2 = 1 + tan^2.

i.e. tan' = 1 + tan^2 implies the deriv of tan^-1 is 1/1+x^2.

[HallsofIvy]

Ooops! You are right. I was thinking integration. There goes me trying to be a smarthmouth again!