Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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?

Answer 5

5

Answer 6

mhm that's correct so that goes in the front

and then afterwards you just substitute y=5x in the fraction

Answer 7

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

Answer 8

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

Answer 9

25x^2

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

can I get a medal ? ^^

Answer 15

thank you :D

Answer 16

thanks for answering my question!

Answer 17

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

Answer 18

x____x oh, yeah it does

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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)...