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

1. jim_thompson5910

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

2. jim_thompson5910

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

3. jim_thompson5910

let me know if that helps or not to get started

4. 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 ?

5. rational

you may memorize but derivation isn't that hard

6. rational

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

7. anonymous

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

8. rational

Exactly!

9. 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

10. rational

ok sure good luck!

11. anonymous

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

12. anonymous

Is that right? I can simplify it more right?

13. anonymous

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

14. amilapsn

please check what you've done for $$sec^2y$$...

15. anonymous

Ohhh I forgot to square it!! Right??

16. amilapsn

hm

17. anonymous

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

18. amilapsn

no

19. anonymous

Oh no no no wait

20. anonymous

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

21. amilapsn

:o)

22. anonymous

Hahaha right?? XD

23. anonymous

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

24. anonymous

Nope I'm still wrong!

25. amilapsn

hm

26. amilapsn

i didn't notice it..

27. anonymous

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

28. amilapsn

hm

29. anonymous

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

30. anonymous

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

31. amilapsn

right.

32. anonymous

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

33. amilapsn

u're welcome.