Question

How can I find the derivative of the following function?
f(x) = 5arctan(xe^x)

Answer 1

if y = arctan(x), then dy/dx = 1/(1+x^2)

Answer 2

you will use the chain rule first, then the product rule will be used later on

Answer 3

let me know if that helps or not to get started

Answer 4

@jim_thompson5910 how do you get 1/(1+x^2) ? Is that just something i have to memorize? For example like how the derivative of sinx = cosx ?

Answer 5

you may memorize but derivation isn't that hard

Answer 6

\[\arctan x = y \implies x = \tan y \implies \frac{dx}{dy}=\sec^2y \]
\[\implies \frac{dy}{dx}=\cos^2y = \frac{1}{1+x^2}\]

Answer 7

Oh so i can use a triangle to find the derivative, correct? Thats what you did above right?

Answer 8

Exactly!

Answer 9

Okay I understand! I'm gonna try to derive this given what I understood, and I'll tell you my answer :D i feel like its gonna be wrong though xD

Answer 10

ok sure good luck!

Answer 11

Ok so this is what I did...

\[\tan(y) = xe^x\]

\[\frac{ dy }{ dx }\tan(y) = \frac{ dy }{ dx } xe^x\]

\[\sec^2 yy' = e^x +xe^x\]

\[y'=\frac{ e^x + xe^x }{ \sec^2 y } = e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\]

\[f'(x) = 5e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\]

Answer 12

Is that right? I can simplify it more right?

Answer 13

I can't think of a way to simplify the denominator

Answer 14

please check what you've done for \(sec^2y\)...

Answer 15

Ohhh I forgot to square it!! Right??

Answer 16

hm

Answer 17

So then it would be

\[y'=5 \frac{ e^x(1+x) }{ 1+2xe ^{2x}+x^2e ^{4x} }\]

Answer 18

no

Answer 19

Oh no no no wait

Answer 20

\[y'=5e^x \frac{ 1+x }{ 1+xe ^{2x} }\]

Answer 21

:o)

Answer 22

Hahaha right?? XD

Answer 23

But then what can I do next? Can I simplify it any more?

Answer 24

Nope I'm still wrong!

Answer 25

hm

Answer 26

i didn't notice it..

Answer 27

The part when I used the Pythagorean theorem to get \[\sqrt{1+xe ^{2x}}\] was wrong! It should be \[\sqrt{1+x^2e ^{2x}}\]

Answer 28

hm

Answer 29

So at the end I get \[y'=5e^x \frac{ 1+x }{ 1+x^2e ^{2x} }\]

Answer 30

Cuz \[(xe^x)^2 = x^2e ^{2x}\] and not \[xe ^{2x} \] is that right?

Answer 31

right.

Answer 32

Awesome!!! Phewww haha that was such a struggle!! Thanks so so much @amilapsn and @rational

Answer 33

u're welcome.