differentiating logs! help!Watch

This discussion is closed.
Thread starter 13 years ago
#1
how do i differentiate this:

ln[(2 + cosx)/(3 - sinx)]

any tips would be great cos my exam is next week!
0
13 years ago
#2
Hint:

ln[(2 + cosx)/(3 - sinx)] = ln[(2 + cosx)] - ln[(3 - sinx)]

then use chain rule.
0
13 years ago
#3
y = ln[(2 + cosx)/(3 - sinx)
y = ln(2 + cosx) - ln(3 - sinx)

ln(f(x)) - ln(g(x))
dy/dx = f'(x)/f(x) - g'(x)/g(x)

so
ln(2 + cosx) - ln(3 - sinx)
= -sinx/(2+cosx) - cosx/(3-sinx)
0
13 years ago
#4
(Original post by Widowmaker)
ln(f(x)) - ln(g(x)) = f'(x)/x - g'(x)/x
well even if the "=" sign denotes differentiation this is still wrong

ln(f(x)) differentiates to f'(x)/f(x) by the chain rule
0
13 years ago
#5
(Original post by RichE)
well even if the "=" sign denotes differentiation this is still wrong

ln(f(x)) differentiates to f'(x)/f(x) by the chain rule
it's been a bad day, "thinks of REM"
0
13 years ago
#6
Okay, to make up for my incompetence.

y = ln(f(x))
u = f(x), therefore y = ln(u)

dy/dx = du/dx x dy/du
dy/dx = f'(x) x 1/u = f'(x)/u

So ln(f(x)) = f'(x)/f(x) since u = f(x)
0
13 years ago
#7
(Original post by Widowmaker)
it's been a bad day, "thinks of REM"
That's a couple of people, so far, today who've been having a bad time.

Must be too much maths in the air.
0
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