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#
