Finding the derivative of f(x)=sin x at f'(1)

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TextClick

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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.
 
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TextClick said:
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.
Click to expand...
You need to use parentheses more carefully:

\(\displaystyle \ f'(1) \ = \ \lim_{h\to 0}\frac{sin(1+h) \ - \ sin(1)}{h}\)

\(\displaystyle = \ \lim_{h\to 0}\frac{sin(1) \ * \ cos(h) \ + \ cos(1) \ * sin(h)\ - \ sin(1)}{h}\)

\(\displaystyle = \ \lim_{h\to 0}\frac{sin(1) \ * \ \left[cos(h) - 1 \ \right] \ }{h} \ + \ cos(1) \ * \ \lim_{h\to 0}\frac{sin(h)\ }{h}\)

continue......
 
Thank you for showing your work.

1st of all the difference quotient is NOT f(x+h)-f(x)/h, rather it is [f(x+h)-f(x)]/h.

You got to f'(1) (not f'(x)) = lim h>0 {sin(1)cos(h)+cos(1)sin(h)-sin(1)} / h. Very good.
Now write the lim as a sum of two limits where in one lim you factor out sin(1).
 
Subhotosh Khan said:
You need to use parentheses more carefully:

\(\displaystyle \ f'(1) \ = \ \lim_{h\to 0}\frac{sin(1+h) \ - \ sin(1)}{h}\)

\(\displaystyle = \ \lim_{h\to 0}\frac{sin(1) \ * \ cos(h) \ + \ cos(1) \ * sin(h)\ - \ sin(1)}{h}\)

\(\displaystyle = \ \lim_{h\to 0}\frac{sin(1) \ * \ \left[cos(h) - 1 \ \right] \ }{h} \ + \ cos(1) \ * \ \lim_{h\to 0}\frac{sin(h)\ }{h}\)

continue......
Click to expand...

Alright, so moving on, I got:

cos (1) * lim h->0 [sin(h)/h] - sin(x) lim h->0 [1-cos(h)/h]

= cos (1)

Is this correct?
 
Jomo said:
Thank you for showing your work.

1st of all the difference quotient is NOT f(x+h)-f(x)/h, rather it is [f(x+h)-f(x)]/h.

You got to f'(1) (not f'(x)) = lim h>0 {sin(1)cos(h)+cos(1)sin(h)-sin(1)} / h. Very good.
Now write the lim as a sum of two limits where in one lim you factor out sin(1).
Click to expand...

Okay, so what I did as follows:

=lim h>0 [cos(1)sin(h)]/h + [sin (1)cos(h)-sin(1)]/h

=lim h>0 cos(1)*[sin(h)/h] + lim h>0 sin(1)*[(cos(h)-1)/h]

=cos(1) lim h>0 sin(h)/h - sin(x) lim h>0 [1-cos(h)]/h

=cos(1)

Did I get the right answer?
 
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