Questions: Use the limit definition of derivatives to show that d/dx cos x = -sin x.

Solution Steps

We start with the limit definition of the derivative for the function \( f(x) = \cos x \): \[ \frac{d}{dx} \cos x = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h} \]

Using the cosine addition formula, we can express \( \cos(x + h) \) as: \[ \cos(x + h) = \cos x \cos h - \sin x \sin h \] Substituting this into the limit expression gives: \[ \frac{d}{dx} \cos x = \lim_{h \to 0} \frac{(\cos x \cos h - \sin x \sin h) - \cos x}{h} \]

We simplify the expression inside the limit: \[ \frac{d}{dx} \cos x = \lim_{h \to 0} \frac{-\sin x \sin h + \cos x (\cos h - 1)}{h} \] This can be separated into two limits: \[ \frac{d}{dx} \cos x = \lim_{h \to 0} \left( -\sin x \frac{\sin h}{h} + \cos x \frac{\cos h - 1}{h} \right) \]

As \( h \to 0 \), we know that: \[ \lim_{h \to 0} \frac{\sin h}{h} = 1 \quad \text{and} \quad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \] Thus, we find: \[ \frac{d}{dx} \cos x = -\sin x \cdot 1 + \cos x \cdot 0 = -\sin x \]

Final Answer