If f(u)=u+1/(cos^2u) , u=g(x)=pi(x), x=1/4 , find the derivative of f(g(x)). I can't seem to get this one, step by step please?

Answer 1

let u = π/x
v = u + 1/(cos^2 u)

OR is it v = (u+1)/(cos^2 u)

I will assume it is the way you typed it, then

v = u + (cos u)^-2
dv/du = 1 - 2(cosu)^-3 (-sinu)
= 1 + 2sinu/cos^3 u

u = πx
when x = 1/4, u = π/4
dv/du = 1 + 2sin(π/4)/cos^3 (π/4)
= 1 + 2tan(π/4) /cos^2 (π/4)
= 1 + 2(1)/(1/2) = 5

du/dx = π

then dv/dx = dv/du * du/dx
= 5 * π
= 5π

better check my work

Answer 2

This looks right! Thanks so much!!

Answer 3

But I was wondering- is it possible to use the (f•g)'= f'(g(X))*g(X)' and still obtain the same answer? For the composite rule, we sub u into the dv/du equation and then take the derivative of that ONCE. Then we multiply that entire function by pi which is the derivative of pix and then sub in the X=1/4. I don't think I did it right and get 5pi. Could you please check the answer again in this way as well please?