Is f'(x) = cos^2(x)/sin(x) for f(x) = ln(sin(x))?

[tod88]

Homework Statement

f(x) = ln(sin(x))

find f'(x)

The Attempt at a Solution

I had a question about this problem. The answer on the website said that f'(x) = cot(x) or cos(x)/sin(x) but I disagreed.

I got d/du ln(u) = du/u so f'(x) = cos(x)/sin(x) * cos(x) << for the chain rule because of the derivative of the "inner" function. So my answer was cos^2(x)/sin(x).

Are they right or am I? Thanks.

 

[D H]

The chain rule is for some function of the form f(g(x)), df/dx = df/dg*dg/gx[/itex]. Here, g(x)=sin(x) and f(g)=ln(g). So, df/dg=1/g=1/sin(x) and dg/dx=cos(x). Putting these results together,

$$\frac d{dx} \ln(\sin x) = \frac 1 {\sin x} \cos x = \frac {\cos x}{\sin x} = \cot x$$

 

[HallsofIvy]

In other words, you applied the chain rule twice when you shouldn't have!

The derivative of ln(sin(x)) is 1/sin(x) times the derivative of sin(x)= 1/sin(x) times cos(x)= cos(x)/sin(x)= cot(x). You had already put the cos(x) in the numerator, then multiplied by it again.

More formally, d(f(u))/dx= df/du * du/dx. Here, u= sin(x) so ln(sin(x))= ln(u) and its derivative is df/du= 1/u= 1/sin(x). Then you multiply by du/dx= cos(x).