Derive cos'x from cos(A)-cos(B)

[Karol]

Homework Statement

Use $~\cos\alpha-\cos\beta~$ to develop the formula for the derivative of cos(x) from the definition:
$$\frac{d(\cos x)}{dx}=\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}$$

Homework Equations

$$\cos\alpha-\cos\beta=(-2)\sin\left( \frac{\alpha+\beta}{2} \right)\cdot \sin\left( \frac{\alpha-\beta}{2} \right)$$
$$\lim\frac{\sin x}{x}=1$$

The Attempt at a Solution

$$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=...=(-2)\lim\frac{\sin\left( \frac{2x+\Delta x}{2} \right)}{\Delta x}\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
Even if i could use $~\lim\frac{\sin x}{X}=1~$, which would eliminate the second member, i can't deal with the first member.

 

[ehild]

Homework Statement

Use $~\cos\alpha-\cos\beta~$ to develop the formula for the derivative of cos(x) from the definition:
$$\frac{d(\cos x)}{dx}=\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}$$

Homework Equations

$$\cos\alpha-\cos\beta=(-2)\sin\left( \frac{\alpha+\beta}{2} \right)\cdot \sin\left( \frac{\alpha-\beta}{2} \right)$$
$$\lim\frac{\sin x}{x}=1$$

The Attempt at a Solution

$$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=...=(-2)\lim\frac{\sin\left( \frac{2x+\Delta x}{2} \right)}{\Delta x}\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
Even if i could use $~\lim\frac{\sin x}{X}=1~$, which would eliminate the second member, i can't deal with the first member.

You divide by Δx too many times.

 

[Karol]

$$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=(-2)\lim\frac{\sin\left( \frac{x+\Delta x+x}{2}\right)\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
$$=(-2)\lim \sin\left( \frac{2x+\Delta x}{2} \right)\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
$$u=\frac{\Delta x}{2}, ~~\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}=\lim\frac{\sin u}{2u}=\frac{1}{2}$$
$$\cos'x=-\sin x$$

 

[ehild]

$$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=(-2)\lim\frac{\sin\left( \frac{x+\Delta x+x}{2}\right)\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
$$=(-2)\lim \sin\left( \frac{2x+\Delta x}{2} \right)\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
$$u=\frac{\Delta x}{2}, ~~\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}=\lim\frac{\sin u}{2u}=\frac{1}{2}$$
$$\cos'x=-\sin x$$

OK.