Find the derivative of $\sin\left(x\right)$ using the definition

Find the derivative of $\sin\left(x\right)$ 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 $\sin\left(x\right)$. Substituting $f(x+h)$ and $f(x)$ on the limit, we get

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

Factoring by $\sin\left(x\right)$

Expand the fraction $\frac{\sin\left(x\right)\left(\cos\left(h\right)-1\right)+\cos\left(x\right)\sin\left(h\right)}{h}$ 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: $\displaystyle\lim_{x\to c}(f(x)\pm g(x))=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x))$

Applying the Sandwich Theorem, which states that: Let $I$ be an interval that contains the point $c$, and let $f(x)$, $g(x)$, and $h(x)$ be functions defined on $I$. If for every $x$ not equal to $c$ in the interval $I$ we have $g(x)\leq f(x)\leq h(x)$ and also suppose that: $\displaystyle\lim_{x\to c}{g(x)}=\lim_{x\to c}{h(x)}=L$, then: $\displaystyle\lim_{x\to c}{f(x)}=L$

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