d/dx(sinxtanx)= ?

Question

Shouldn't the derivative of this be (cosxtanx + sinxsec^2x)? My book says (cosxtanx+secxtanx). Why!?!? Am I missing a trig identity or something? I'm trying to learn calculus and it doesn't explain why at all! Please help and thanks in advance!

Answer 1

sin x sec² x
= sin x sec x sec x
= sin x (1 / cos x) sec x
= (sin x / cos x) sec x
= tan x sec x

you got it right :) I don't know why the book wrote the answer this way....

Answer 2

You are right as explained by a^2 + b^2 = c^2. Book has written the answer after simplification. Normally it is expected that the answer be written in simplified form.

Answer 3

using the product rule,
d(sin x tan x)/dx =
sin x d(tan x)/dx + tan x d(sin x)/dx =
sin x sec² x + tan x cos x
as you say.

but sin x sec² x = sin x (1/cos² x) =
(sin x / cos x)(1/cos x) =
tan x sec x
as the book says.

also tan x cos x + sin x sec² x =
(sin x / cos x)(cos x) + (sin x)(1/cos² x) =
sin x + sin x / cos²x =
sin x (1 + 1/cos² x) =
[sin x (cos² x + 1)] / cos² x
as my ti-89 says

Answer 4

d/dx(sinxtanx) = sinx*d/dx(tanx) + tanx*d/dx(sinx)

=>sinx(sec^2(x) + tanx(cosx)

=>(sinx/cosx)*(secx) + tanx cosx = tanxsecx + tanxcosx

both answers are the same