- anonymous

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

- schrodinger

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- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

let me know if that helps or not to get started

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## More answers

- anonymous

@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 ?

- rational

you may memorize but derivation isn't that hard

- rational

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

- anonymous

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

- rational

Exactly!

- anonymous

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

- rational

ok sure good luck!

- anonymous

Ok so this is what I did...
\[\tan(y) = xe^x\]
\[\frac{ dy }{ dx }\tan(y) = \frac{ dy }{ dx } xe^x\]
\[\sec^2 yy' = e^x +xe^x\]
\[y'=\frac{ e^x + xe^x }{ \sec^2 y } = e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\]
\[f'(x) = 5e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\]

- anonymous

Is that right? I can simplify it more right?

- anonymous

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

- amilapsn

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

- anonymous

Ohhh I forgot to square it!! Right??

- amilapsn

hm

- anonymous

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

- amilapsn

no

- anonymous

Oh no no no wait

- anonymous

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

- amilapsn

:o)

- anonymous

Hahaha right?? XD

- anonymous

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

- anonymous

Nope I'm still wrong!

- amilapsn

hm

- amilapsn

i didn't notice it..

- anonymous

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

- amilapsn

hm

- anonymous

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

- anonymous

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

- amilapsn

right.

- amilapsn

u're welcome.

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