Find the derivative of $\cos\left(2x\right)$ using the definition. Apply the definition of the derivative: $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. The function $f(x)$ is the function we want to differentiate, which is $\cos\left(2x\right)$. Substituting $f(x+h)$ and $f(x)$ on the limit, we get

Multiply the single term $2$ by each term of the polynomial $\left(x+h\right)$

Using the cosine of a sum formula: $\cos(\alpha\pm\beta)=\cos(\alpha)\cos(\beta)\mp\sin(\alpha)\sin(\beta)$, where angle $\alpha$ equals $2x$, and angle $\beta$ equals $2h$

Factoring by $\cos\left(2x\right)$

Expand the fraction $\frac{\cos\left(2x\right)\left(\cos\left(2h\right)-1\right)-\sin\left(2x\right)\sin\left(2h\right)}{h}$ into $2$ simpler fractions with common denominator $h$

The limit of a sum of two or more functions is equal to the sum of the limits of each function: $\displaystyle\lim_{x\to c}(f(x)\pm g(x))=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x))$

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

Knowing that $\displaystyle\lim_{h\to 0}{\left(\frac{\cos(h)-1}{h}\right)}=0$

Any expression multiplied by $0$ is equal to $0$

$x+0=x$, where $x$ is any expression

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

The limit of the product of a function and a constant is equal to the limit of the function, times the constant: $\displaystyle \lim_{t\to 0}{\left(at\right)}=a\cdot\lim_{t\to 0}{\left(t\right)}$

Apply the formula: $\lim_{h\to0}\left(\frac{\sin\left(nh\right)}{h}\right)$$=n$, where $n=2$