Question

How do you differentiate `f(x)=ln(sinx)/cosx` using the quotient rule?

Answer 1

`f'(x)=(cosxcotx+sinxln(sinx))/cos^2x`

Explanation:

The quotient rule states that

`d/dx[g(x)/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2`

When this is applied to `f(x)`, the function at hand, we see that

`f'(x)=(cosxd/dx[ln(sinx)]-ln(sinx)d/dx[cosx])/cos^2x`

Find each derivative separately:

To differentiate the natural logarithm function, use the chain rule, which for a natural logarithm function states that

`d/dx[ln(k(x))]=1/(k(x))*k'(x)`

Here, `k(x)=sinx`, so

`d/dx[ln(sinx)]=1/sinx*d/dx[sinx]=1/sinx*cosx=cotx`

As for the other derivative,

`d/dx[cosx]=-sinx`

Plugging these both back in, we see that

`f'(x)=(cosx(cotx)-ln(sinx)(-sinx))/cos^2x`

This can be written as

`f'(x)=(cosxcotx+sinxln(sinx))/cos^2x`

As with many trigonometric functions, this can be rewritten in many ways, including

`f'(x)=secx(cotx+tanxln(sinx))`