Find the derivative of f when f(x)= (sin x) / (5-4cosx)?

Question

A) f'(x)= (5cosx+4) / ((5-4cosx)^2)
B) f'(x)= (5sinx -4cosx^2 x) / ((5-4cos x)^2)
C) f'(x)= (5cosx+4) / (5-4cos x)
D) f'(x)= (5cosx - 4sin^2x) / ((5-4cosx)^2)
E) f'(x)= (5cosx -4) / ((5 -4cosx)^2)
F) f'(x)= (5sinx - 4) / (5 - 4cosx)

Answer 1

f(x) = sin x / (5 - 4cos x)

Using quotient rule ;-

f `(x) = [(5 - 4 cos x) cos x - sinx. 4sinx] / (5 - 4cos x)²

f `(x) = [5cos x - 4(cos²x + sin²x)] / (5 - 4 cos²x)

f ` (x) = [ 5cos x - 4] / (5 - 4 cos²x)
This is answer E

Answer 2

Use the quotient rule for taking derivatives.

E) f'(x)= (5cosx - 4) / ((5 - 4cosx)^2)

Answer 3

let f(x)=sinx , g(x)=(5-4cosx) then using quotient rule

{(5-4cosx)cosx - sinx(0-4(-sinx))}/(5-4cosx)^2

{5cosx-4cos^2x - 4sin^2x}/(5-4cosx)^2
but cos^2x+sin^2x=1
{5cosx - 4(cos^2x+sin^2x)}/(5-4cosx)^2

{5cosx - 4}/(5-4cosx)^2

which is option E

Answer 4

ln(y) = 5 sin(x) ln(x) Take derivatives of both elements (using the chain rule on the left, the product rule on the important ideal): a million/y dy/dx = 5 cos(x) ln(x) + 5 sin(x) / x Multiply through ability of y: dy/dx = y( 5 cos(x) ln(x) + 5 sin(x) / x ) or dy/dx = x^(5 sin(x)) ( 5 cos(x) ln(x) + 5 sin(x) / x )

Answer 5

use quotient rule which works out b e! good luck

Answer 6

Do your own homework. If your having problems---ask your teacher