Can someone show me how to find
d/dx (arctan (5x))

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- anonymous

Can someone show me how to find
d/dx (arctan (5x))

- katieb

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- UsukiDoll

\[\frac{d}{dx} \tan^{-1}(5x) \] this?

- UsukiDoll

ok. the formula for the derivative of the inverse trigonometric function is \[\tan^{-1}(x) = \frac{1}{1+x^2}\]

- UsukiDoll

so we need to find the derivative of 5x first
what is the derivative of 5x? @cstarxq

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## More answers

- UsukiDoll

I'm not using the right variable here...this is going to be confusing a bit
\[\frac{d}{dy} \]\[\tan^{-1}(y) = \frac{1}{1+y^2}\]
let y = 5x
so we need the derivative of 5x which is?

- anonymous

5

- UsukiDoll

mhm that's correct so that goes in the front
and then afterwards you just substitute y=5x in the fraction

- UsukiDoll

\[\frac{5}{1+y^2}\]
so if y =5x
then we have (5x)^2
which is

- UsukiDoll

(5x)^2 = (5x)(5x) = ?

- anonymous

25x^2

- UsukiDoll

correct and that's it..
\[\frac{5}{1+25x^2}\]

- UsukiDoll

you just have to take the derivative of the inside for \[\frac{d}{dx} \tan^{-1}(5x) \]
in that case it was a 5x so the derivative of that is 5

- anonymous

oh i see, i forgot i had to take the derivative of 5x and substituting that into y^2..

- UsukiDoll

then the derivative goes on the numerator.. and then the denominator is just pure substitution

- UsukiDoll

can I get a medal ? ^^

- UsukiDoll

thank you :D

- anonymous

thanks for answering my question!

- UsukiDoll

you're welcome :)
If I remembered correctly, you need to memorize them >.<
Calculus II had so much formulas.

- anonymous

x____x oh, yeah it does

- UsukiDoll

Calculus III should be easy, but Calculus IV... UGH! So many conversations, integrations, and vectors come back

- anonymous

I lost my connection, but I'm taking calc III and linear algebra next semester. Hopefully it won't be too bad.

- UsukiDoll

I've already taken those subjects. It's not too bad

- idku

whether you are still viewing or not, there is a way of doing it. Don't be a slave of a non-making sense (or even making sense) formula. Every formula you use should be something you understand and that is what I will try to improve right now.

- idku

you are dealing with an inverse function. Let's say inv. tan.
\(y={\rm Tan}^{-1}(x)\)
re-write it first.
\(x={\rm Tan}^{}(y)\)
(i hope you know why I did this, or elsego back and rvw that rule)
Now differentiate with respect to x.
(Taking a derivative of a function y=g(x), that is dy/dx)
\(\frac{dy}{dx}~x=\frac{dy}{dx}~{\rm Tan}^{}(y)\)
\(1={\rm Sec}^{2}(y)\times y'\)
\(1/{\rm Sec}^{2}(y)= y'\)
\(1/\left({\rm Tan}^{2}(y)+1\right) = y'\)
you know what tan(y) is from the beginning, and thus you know what tan^2(y) is as well.
\(1/\left(x^2+1\right) = y'\)
and there is your derivative

- idku

same you can do with any inverse function.....

- anonymous

Yes, I'm still reviewing this. Thanks @idku, I just used your method to evaluate the derivative for the original problem I was trying to figure out which was actually using integration by parts to evaluate integral of arctan(5x)...

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