I'll assume you mean ln(sinx)
First principles of derivates says that for f(x), f'(x)=lim_(h->0)(f(x+h)-f(x))/h
So, we shall apply it to the function given.
f(x)=ln(sinx)
f'(x)=lim_(h->0)(ln(sin(x+h))-ln(sinx))/h
Using the addition formula, we get sin(A+B)=sinAcosB+cosAsinB
f'(x)=lim_(h->0)(ln(sin(x)cos(h)+cos(x)sin(h))-ln(sinx))/h
We can also use the subtraction that says that: log_a(b)-log_a(c)=log_a(b/c) to get:
f'(x)=lim_(h->0)(ln((sin(x)cos(h)+cos(x)sin(h))/sinx))/h
color(white)(f'(x))=lim_(h->0)(ln((sin(x)cos(h))/sinx+(cos(x)sin(h))/sinx))/h
color(white)(f'(x))=lim_(h->0)(ln(cos(h)+cot(x)sin(h)))/h
color(white)(f'(x))=lim_(h->0)ln((cos(h)+cot(x)sin(h))^(1/h))
color(white)(f'(x))=lim_(h->0)ln(cos(h)^(1/h)*(1+cot(x)tan(h))^(1/h))
color(white)(f'(x))=ln(lim_(h->0)cos(h)^(1/h)*lim_(h->0)(1+cot(x)tan(h))^(1/h))
lim_(h->0)(1+h)^(1/h)=e
So:
lim_(h->0)(1+f(h))^(1/h)=e
And:
lim_(h->0)(1+af(h))^(1/h)=e^(a)
f'(x)=ln(lim_(h->0)cos(h)^(1/h)*e^cotx)
lim_(h->0)cos(h)^(1/h)=1
Proof:
As h tends to 0, cos(h) tends to 1. When h<1, cos(h) gets raised to the power of a very large number. So we end up with effectively 1^oo=1
f'(x)=(1*e^cotx)
color(white)(f'(x))=ln(e^cotx)
color(white)(f'(x))=cotx
