a is a constant here which is throwing me off
2007-01-10 16:32:05 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x) = sin(x+ a cosx)
f'(x) = cos( x +acosx) [1 -a sinx] .
2007-01-10 17:18:04 · answer #1 · answered by Anonymous · 4⤊ 0⤋
chain rule- derivative of outside multipled by derivative of whats inside the brackets:
= cos ( x + a cos x) * dy/dx of (x+ a cos x)
= cos ( x + a cos x) * (1 - a sinx)
similarly : other chain rule method: dy/dx = dy/du * du/dx:
let u = x + a cos x ,
du/dx= 1 - a sin x
y= sin (x + a cos x) = sin (u)
dy/du = cos (u)
therefore
dy/dx = dy/du * du/dx
= cos (u) * ( 1- a sin x)
=cos (x + a cos x) * ( 1 - a sin x)
2007-01-10 16:58:59 · answer #2 · answered by anonymous 2 · 1⤊ 0⤋
cos(x+a cosx) * (1+a(-sinx))
cos(x+a cosx) * (1- a sinx))
by chain rule
2007-01-10 16:35:36 · answer #3 · answered by Professor Maddie 4 · 3⤊ 0⤋
cos( x +acosx) (1 -a sinx)
2007-01-11 01:58:23 · answer #4 · answered by lobis3 5 · 3⤊ 0⤋
d{sin(x + a cos x)}/dx
= (1 - a sin x)(cos(x + a cos x))
2007-01-10 20:09:34 · answer #5 · answered by Northstar 7 · 1⤊ 1⤋
divide by x
2007-01-10 16:35:58 · answer #6 · answered by megmotox 3 · 0⤊ 3⤋