MHB How to find the value of f'(x) at c

[rahulk1]

How to find the value of f'(x) at cif

1. f(x)=sinx, c=pi/4
2. f(x)=sinx, c=3pi/2
3.2. f(x)=cosx, c=0, pi/2, pi, 3pi/2, 2pi

 

[MarkFL]

Since you posted in the Pre-Calculus forum, I suppose you are to use the definition:

$$f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

So, for $f(x)=\sin(x)$, we have:

$$f'(x)\equiv\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}$$

Using the angle-sum identity for the sine function, we may write:

$$f'(x)=\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using a Pythagorean identity on the first limit on the RHS, we obtain:

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin^2(h)}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin(h)}{h}\cdot\lim_{h\to0}\frac{\sin(h)}{(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using the fact that $$\lim_{u\to0}\frac{\sin(u)}{u}=1$$, we have:

$$f'(x)=-\sin(x)(1)(0)+\cos(x)(1)=\cos(x)$$

Can you now obtain a result for $f(x)=\cos(x)$?

 

[rahulk1]

Since you posted in the Pre-Calculus forum, I suppose you are to use the definition:

$$f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

So, for $f(x)=\sin(x)$, we have:

$$f'(x)\equiv\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}$$

Using the angle-sum identity for the sine function, we may write:

$$f'(x)=\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using a Pythagorean identity on the first limit on the RHS, we obtain:

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin^2(h)}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin(h)}{h}\cdot\lim_{h\to0}\frac{\sin(h)}{(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using the fact that $$\lim_{u\to0}\frac{\sin(u)}{u}=1$$, we have:

$$f'(x)=-\sin(x)(1)(0)+\cos(x)(1)=\cos(x)$$

Can you now obtain a result for $f(x)=\cos(x)$?

Hi

Thanks for your suggestion

But what about c=pi/4 where are used

c=pi/4 in the answer

 

[MarkFL]

Hi

Thanks for your suggestion

But what about c=pi/4 where are used

c=pi/4 in the answer

Well, if $f'(c)=\cos(c)$ and $$c=\frac{\pi}{4}$$...then what do you get?

 

[rahulk1]

Since you posted in the Pre-Calculus forum, I suppose you are to use the definition:

$$f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

So, for $f(x)=\sin(x)$, we have:

$$f'(x)\equiv\lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}$$

Using the angle-sum identity for the sine function, we may write:

$$f'(x)=\lim_{h\to0}\frac{\sin(x)(\cos(h)-1)+\cos(x)\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=\sin(x)\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using a Pythagorean identity on the first limit on the RHS, we obtain:

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin^2(h)}{h(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

$$f'(x)=-\sin(x)\lim_{h\to0}\frac{\sin(h)}{h}\cdot\lim_{h\to0}\frac{\sin(h)}{(\cos(h)+1)}+\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$

Using the fact that $$\lim_{u\to0}\frac{\sin(u)}{u}=1$$, we have:

$$f'(x)=-\sin(x)(1)(0)+\cos(x)(1)=\cos(x)$$

Can you now obtain a result for $f(x)=\cos(x)$?

How sin(h)/(cos(h)+1 =0 please descibe

 

[MarkFL]

How sin(h)/(cos(h)+1 =0 please descibe

$$\lim_{h\to0}\frac{\sin(h)}{\cos(h)+1}=\frac{\sin(0)}{\cos(0)+1}=\frac{0}{1+1}=0$$ :D