Derivative of cos square x by First Principle | cos^2x Derivative

The function cos square x is the product of two cosine functions. The derivative of cos square x is equal to -sin2x. In this post, we will find the derivative of cos square x.

Derivative of cos^2 x by First Principle

The first principle of derivatives says that the derivative of a function f(x) is given by the following limit:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

Step 1: In the above formula, we put $f(x)=\cos^2x$. So the derivative of cos square x is equal to

$\dfrac{d}{dx}(\cos^2x)$ $= \lim\limits_{h \to 0} \dfrac{\cos^2(x+h)-\cos^2x}{h}$

Step 2: Now, we will apply the formula of $\cos(x+h)$ $=  \cos x \cos h -\sin x \sin h$.

$\dfrac{d}{dx}(\cos^2x)$ $= \lim\limits_{h \to 0}$ $[( \cos x \cos h -\sin x \sin h)^2 – \cos^2x]/h$

$= \lim\limits_{h \to 0}$  $(\cos^2 x \cos^2 h -2\cos x \cos h\sin x \sin h$ $+\sin^2x\sin^2h – \cos^2x)$/h,  as we know that $(a-b)^2$ $=a^2-2ab+b^2$

$= \lim\limits_{h \to 0}$  $[\dfrac{\cos^2 x (\cos^2 h-1)}{h}$ $- \dfrac{2\cos x \cos h\sin x \sin h}{h}$ $+\dfrac{\sin^2x\sin^2h}{h}]$

Step 3: Now, we apply the Sum rule of limits.

$= \lim\limits_{h \to 0}$  $\dfrac{\cos^2 x (\cos^2 h-1)}{h}$ $- \lim\limits_{h \to 0}$ $\dfrac{2\cos x \cos h\sin x \sin h}{h}$ $+ \lim\limits_{h \to 0}$ $\dfrac{\sin^2x\sin^2h}{h}$

$=\cos^2 x \lim\limits_{h \to 0}$  $\dfrac{(\cos h-1)(\cos h+1)}{h}$ $-2 \lim\limits_{h \to 0}$ $\left(\cos x \cos h\sin x\dfrac{\sin h}{h} \right)$ $+ \lim\limits_{h \to 0}$ $\left( \sin^2 x \sin h \dfrac{\sin h}{h} \right)$

Step 4: Now, we will apply the following rule of limit: the limit of sinh/h is 1 when h tends to 0. Thus, the above is

$=\cos^2 x (\cos 0+1) \lim\limits_{h \to 0}$  $\dfrac{\cos h-1}{h}$ $-2\cos x \cos 0\sin x \cdot 1$ $+ \sin^2 x \sin 0 \cdot 1$

$=2 \cos^2 x \lim\limits_{h \to 0}$  $\dfrac{\cos h-1}{h}$ $-2\sin x\cos x$ $+ 0$ as we know that cos0 =1 and sin0=0.

Step 5: By the L’Hospital rule, we have that the limit $\lim\limits_{h \to 0}$  $\dfrac{\cos h-1}{h}$ $=\lim\limits_{h \to 0} \dfrac{-\sin h}{1}$ $=0$.

So the above quantity is equal to

$2 \cos^2 x \cdot 0$ $-2\sin x\cos x$ $+0$

$=0-2\sin x\cos x+ 0$

$=-2\sin x \cos x$

So the derivative of cos square x is equal to -2sin x cos x and this is obtained by the first principle of derivatives.

Remark: Applying the trigonometric formula of sin 2x=2sin x cos x, we obtain that the derivative of cos^2x is -sin2x.

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