Question

How do you differentiate F(x) = sin(2x)*ln(sec x )?
( -1/2 PI less than X less than 1/2 PI )
Please help !!! I'm really stuck !!

Answer 1

take it as d[f(x)*g(x)]/dx = [df(x)/dx]g(x) + f(x)[dg(x)/dx]

Answer 2

F'(x)= 2cos(2x)*ln(secx) + [sin(2x)/secx]*secx*tanx

Answer 3

you do these things in steps. First step is use the product rule.
d (u v) = u v' + v u'

how far did you get ?

Answer 4

I will write down what I have done. I did use the product rule first, then the composite rule, but I think I have done something wrong

Answer 5

So I used product rule :
1. f ' X= 2 cos (2x) * ln (sec x) + sin (2x) * d\dx ln ( sec x )

2. f ' x = 2 cos (2x) * ln ( sec x ) + sin ( 2x ) * ( 1\ sec x ) ( sec x tan x )- composite rule I used here

3. f ' x =2 cos (2x) * ( ln ( sec x ) ) + sin ( 2x ) ( (sec x tan x) \ sec x )

4. f'x = 2 cos (2x ) * ln (1\ cos x ) + sin (2x) tan x

Answer 6

yes, that looks good. we can use the trig identity sin(2x) = 2 sin(x) cos(x)
and tan(x)= sin(x)/cos(x) to simplify it a bit more:
sin (2x) tan x= 2 sin(x) cos(x) * sin(x)/cos(x) = 2 sin^2(x)

and we have
2 cos (2x ) * ln (1\ cos x ) + 2 sin^2(x)

You are given limits -pi/2 to +pi/2

so you evaluate that between the lower and upper limits

Answer 7

Thank you so much !! I understand now how you simplified more, would you please show how you evaluate between those lower and upper limits

Answer 8

Oh yes, thank you ! I remember now ! Thank you very much !!!