anonymous
  • anonymous
How can I find the derivative of the following function? f(x) = 5arctan(xe^x)
Mathematics
schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
if y = arctan(x), then dy/dx = 1/(1+x^2)
jim_thompson5910
  • jim_thompson5910
you will use the chain rule first, then the product rule will be used later on
jim_thompson5910
  • jim_thompson5910
let me know if that helps or not to get started

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anonymous
  • 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
  • rational
you may memorize but derivation isn't that hard
rational
  • 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
  • anonymous
Oh so i can use a triangle to find the derivative, correct? Thats what you did above right?
rational
  • rational
Exactly!
anonymous
  • 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
  • rational
ok sure good luck!
anonymous
  • 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
  • anonymous
Is that right? I can simplify it more right?
anonymous
  • anonymous
I can't think of a way to simplify the denominator
amilapsn
  • amilapsn
please check what you've done for \(sec^2y\)...
anonymous
  • anonymous
Ohhh I forgot to square it!! Right??
amilapsn
  • amilapsn
hm
anonymous
  • anonymous
So then it would be \[y'=5 \frac{ e^x(1+x) }{ 1+2xe ^{2x}+x^2e ^{4x} }\]
amilapsn
  • amilapsn
no
anonymous
  • anonymous
Oh no no no wait
anonymous
  • anonymous
\[y'=5e^x \frac{ 1+x }{ 1+xe ^{2x} }\]
amilapsn
  • amilapsn
:o)
anonymous
  • anonymous
Hahaha right?? XD
anonymous
  • anonymous
But then what can I do next? Can I simplify it any more?
anonymous
  • anonymous
Nope I'm still wrong!
amilapsn
  • amilapsn
hm
amilapsn
  • amilapsn
i didn't notice it..
anonymous
  • 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
  • amilapsn
hm
anonymous
  • anonymous
So at the end I get \[y'=5e^x \frac{ 1+x }{ 1+x^2e ^{2x} }\]
anonymous
  • anonymous
Cuz \[(xe^x)^2 = x^2e ^{2x}\] and not \[xe ^{2x} \] is that right?
amilapsn
  • amilapsn
right.
anonymous
  • anonymous
Awesome!!! Phewww haha that was such a struggle!! Thanks so so much @amilapsn and @rational
amilapsn
  • amilapsn
u're welcome.

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