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

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if y = arctan(x), then dy/dx = 1/(1+x^2)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3you will use the chain rule first, then the product rule will be used later on

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3let me know if that helps or not to get started

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2you may memorize but derivation isn't that hard

rational
 one year ago
Best ResponseYou've already chosen the best response.2\[\arctan x = y \implies x = \tan y \implies \frac{dx}{dy}=\sec^2y \] \[\implies \frac{dy}{dx}=\cos^2y = \frac{1}{1+x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so i can use a triangle to find the derivative, correct? Thats what you did above right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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
 one year ago
Best ResponseYou've already chosen the best response.0Is that right? I can simplify it more right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't think of a way to simplify the denominator

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0please check what you've done for \(sec^2y\)...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh I forgot to square it!! Right??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then it would be \[y'=5 \frac{ e^x(1+x) }{ 1+2xe ^{2x}+x^2e ^{4x} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y'=5e^x \frac{ 1+x }{ 1+xe ^{2x} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But then what can I do next? Can I simplify it any more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope I'm still wrong!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The part when I used the Pythagorean theorem to get \[\sqrt{1+xe ^{2x}}\] was wrong! It should be \[\sqrt{1+x^2e ^{2x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So at the end I get \[y'=5e^x \frac{ 1+x }{ 1+x^2e ^{2x} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cuz \[(xe^x)^2 = x^2e ^{2x}\] and not \[xe ^{2x} \] is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome!!! Phewww haha that was such a struggle!! Thanks so so much @amilapsn and @rational
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