So the problem I got is as follows. I have to find the derivative of f(x)=sin x at f'(1) x =1 ........................[edited].
So I began by using the limit definition f'(x)=lim h>0 f(x+h)-f(x)/h.
I plugged in 1 into the limit definition to get f'(1)=lim h>0 f(1+h)-f(1)/h.
I then solved for f(1) and f(1+h). For f(1) I got sin (1), and for f(1+h) I got sin (1+h).
I then used the trig identity sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to expand sin (1+h).
So it became sin(1)cos(h)+cos(1)sin(h).
I plugged both values into the limit definition.
I got f'(x)=lim h>0 sin(1)cos(h)+cos(1)sin(h)-sin(1) / h.
After that point, I was stuck and did not know how to continue.
So I began by using the limit definition f'(x)=lim h>0 f(x+h)-f(x)/h.
I plugged in 1 into the limit definition to get f'(1)=lim h>0 f(1+h)-f(1)/h.
I then solved for f(1) and f(1+h). For f(1) I got sin (1), and for f(1+h) I got sin (1+h).
I then used the trig identity sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to expand sin (1+h).
So it became sin(1)cos(h)+cos(1)sin(h).
I plugged both values into the limit definition.
I got f'(x)=lim h>0 sin(1)cos(h)+cos(1)sin(h)-sin(1) / h.
After that point, I was stuck and did not know how to continue.
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