differentiating logs! help!

K2theJ

how do i differentiate this:

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

any tips would be great cos my exam is next week!

Reply 1

Fermat

Hint:

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

then use chain rule.

Reply 2

Christophicus

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)

Reply 3

RichE

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

Reply 4

Christophicus

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" :frown:

Reply 5

Christophicus

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)