Question

How would you differentiate f(x)= [ e^arctanx] \ 1+x^2?
I used the Quotient rule as u= e^arctanx and v=1+x^2 so, I have

f ' (x)= [ (1+x^2) d\du e ^arctanx - e^arctanx * d\dv(1+x^2) ] \ (1+ x^2)^2
Then I used the composite rule ,
with g(u)= e^u g'(u)= e^u

u= f(x) = arctan x f'(x)= 1\ 1+x^2

so f'(x) = [ (1+x^2)*(e^arctanx)*(1\1+x^2)- e^arctanx*(2x) ] \ (1+x^2)^2

f'(x)=[ e^arctanx - e^arctanx * (2x) ] \ (1+x^2 )^2
f'(x)= -2x \ (1+x^2 )
Is this ok? Please help, many thanks !

Answer 1

im thinking logs ...

f(x)= [ e^arctanx] \ 1+x^2

log(f) = log ([ e^arctanx] \ 1+x^2)

log(f) = log (e^arctanx) - log(1+x^2)

log(f) = arctanx - log(1+x^2)

f'/f = (arctanx)' - 2x/(1+x^2)

f' = f [(arctanx)' - 2x/(1+x^2)]

Answer 2

arctan derivative is what ... 1/1+x^2 if memory serves?

Answer 3

im thinking then ....

e^(arctanx) * (1-2x)
f' = -----------------
(1+x^2) ^2

Answer 4

i think you lost track of the quotient rule along the way, which is easy to do since it tends to be the most convoluted rule

Answer 5

with g(u)= e^u g'(u)= e^u
^^^^^^^
chain rule u' e^u

Answer 6

Thank you !!! :)