Question

Use the definition of the derivative, a trigonometric identity, and known trigonometric limits to prove that $\frac{d}{dx}(\cos x)=-\sin x$

Short answer

The trigonometric limit $\frac{d}{dx}(\cos x)=-\sin x$ has been proved.

Step-by-step solution

Step 1. Given Information.

The trigonometric identity:

$\frac{d}{dx}(\cos x)=-\sin x$

Step 2. Definition of derivative.

From the definition derivative,

$f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Step 3. Obtain the derivative for the given function.

From the definition of derivative,

$$\begin{gathered} f\text{'}(\cos x)=\underset{h\to 0}{lim}\left[\frac{\cos (x+h)-\cos x}{h}\right] \\ =\underset{h\to 0}{lim}\left[\frac{(\cos x\cos h-\sin x.\sin h)-\cos x}{h}\right] \\ =\underset{h\to 0}{lim}\left[\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\right] \\ =\underset{h\to 0}{lim}[\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}] \end{gathered}$$

Step 4. Apply the limits.

So the function become,

$$\begin{gathered} f\text{'}(\cos x)=\cos x\underset{h\to 0}{lim}\frac{\cos h-1}{h}-\sin x\underset{h\to 0}{lim}\frac{\sin h}{h} \\ =\cos x\left(0\right)-\sin x\left(1\right) \\ =-\sin x \end{gathered}$$