Q.Find the derivative of $\cos^{2}x$ from the first principle.

Apply the definition $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ using the identity $\cos^2A-\cos^2B=-\sin(A+B)\sin(A-B)$.

Let $f(x)=\cos^{2}x$. By the first-principles definition:

$f'(x) = \lim_{h\to 0} \frac{\cos^{2}(x+h) - \cos^{2}x}{h}$

Use the identity $\cos^{2}A - \cos^{2}B = -\sin(A+B)\sin(A-B)$ with $A=x+h$, $B=x$:

$\cos^{2}(x+h) - \cos^{2}x = -\sin(2x+h)\sin(h)$

So:

$f'(x) = \lim_{h\to 0} \frac{-\sin(2x+h)\sin h}{h} = -\lim_{h\to0}\sin(2x+h) \cdot \lim_{h\to0}\frac{\sin h}{h}$

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