Derivatives..?

Question

Find the derivatives of the function
f(x) = [(sin x)^2(tan x)^2 ]/ (x^2 + 1)^2

Answer 1

f(x) = (sin(x)^2 * tan(x)^2)/((x^2 + 1)^2)

f'(x) = (((x^2 + 1)^2 * (sin(x)^2 * tan(x)^2)') - (((x^2 + 1)^2)' * (sin(x)^2 * tan(x)^2))/((x^2 + 1)^3)

(sin(x)^2 + tan(x)^2)' = (sin(x)^2' * tan(x)^2) + (sin(x)^2 * tan(x)^2') = (2sin(x)cos(x) * tan(x)^2) + (sin(x)^2 * 2sec(x)^2tan(x)) =
(2sin(x)cos(x) * (sin(x)/cos(x))^2) + (2sin(x)^2 * (1/cos(x))^2 * tan(x)) = (2(sin(x)/cos(x))) + (2(sin(x)/cos(x))^2 * tan(x)^2) = 2tan(x) + (2tan(x)^2 * tan(x)^2) = 2tan(x)^2 + 2tan(x)^4

f'(x) = (((x^2 + 1)^2 * (2tan(x)^2 + 2tan(x)^4) - ((x^2 + 1)^2' * sin(x)^2 * tan(x)^2)) / ((x^2 + 1)^3)

f'(x) = (((x^2 + 1)^2 * (2tan(x)^2)(1 + tan(x)^2)) - (((x^2 + 1)(x^2 + 1))' * sin(x)^2 * tan(x)^2)) / ((x^2 + 1)^3)

f'(x) = (((2tan(x)^2)(x^2 + 1)^2 * sec(x)^2) - ((x^4 + 2x^2 + 1)' * sin(x)^2 * tan(x)^2)) / ((x^2 + 1)^3)

f'(x) = (((2tan(x)^2)((x^2 + 1)sec(x))^2) - ((4x^3 + 4x)(sin(x)tan(x))^2)) / ((x^2 + 1)^3)

f'(x) = (((2tan(x)^2)((x^2 + 1)sec(x))^2) - 4x((x^2 + 1)(sin(x)tan(x))^2)) / ((x^2 + 1)^3)

f'(x) = (((2tan(x)^2)(x^2 + 1)sec(x)^2) - 4x(sin(x)tan(x))^2)/((x^2 + 1)^2)

ANS :
f'(x) = (((2tan(x)^2)(x^2 + 1)sec(x)^2) - 4x(sin(x)tan(x))^2)/((x^2 + 1)^2)

or

f'(x) = ((2tan(x)^2)/(x^2 + 1)^2)((x^2 + 1)sec(x)^2 - 2xsin(x))

I probably did something wrong somewhere, so i am not 100% on the answer.

Answer 2

just do your homework, this is a long derivative that'd be bad to work out on yahoo. Just do the quotient rule[(f'g - fg')/g^2] and the exponent rule [n(u^(n-1))(u')

Answer 3

i agree, this one is too big and too complicated. use the product, chain, and quotient rules for this. for more info see: http://www.mathematicshelpcentral.com/lecture_notes/calculus_1.htm