# Derivative of sinx formula Proof

The differentiation of sin function can be derived in calculus by the principle relation of derivative of function in limit form.

### Express Differentiation of function in Limit form

According to the relation of derivative of a function with respect to $x$ in terms of limits.

$\dfrac{d}{dx} \, f(x) = \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{f(x+h)-f(x)}{h}$

Take $f(x) = \sin{x}$, then $f(x+h) = \sin{(x+h)}$ and substitute them in the limit formula to start deriving the proof of deriving the derivative of sine function with respect to $x$.

$\implies \dfrac{d}{dx} \, \sin{x}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\sin{(x+h)}-\sin{x}}{h}$

### Use sum to product transformation rule

Use sum to product identity of sine to express them in product form and then simplify it.

$= \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{2\cos{\Bigg[\dfrac{x+h+x}{2}\Bigg]}\sin{\Bigg[\dfrac{x+h-x}{2}\Bigg]}}{h}$

$\require{cancel} = \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{\cancel{x}+h-\cancel{x}}{2}\Bigg]}}{h}$

$= \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{2\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}$

### An adjustment for further simplification

The number $2$ multiplies the numerator and it divides the denominator. So, move the number $2$ to denominator.

$= \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}$

$= \displaystyle \large \lim_{h \,\to\, 0} \normalsize \cos{\Bigg[\dfrac{2x+h}{2}\Bigg]} \times \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}$

### Apply Limit Product Rule

Use limit product law to split expression as two multiplying factors.

$= \displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}$

### Simplifying the mathematical expression

The sine function consists $\dfrac{h}{2}$ as angle and the denominator contains same. It represents the lim sinx/x as x approaches 0 formula but the limit is not $\dfrac{h}{2}$ approaches zero. So, change the limit $h$ to $\dfrac{h}{2}$ term. Actually $h$ tends to $0$ ($h \to 0$) then $\dfrac{h}{2}$ also tends to zero ($\dfrac{h}{2} \to 0$). Therefore, replace $h \to 0$ by $\dfrac{h}{2} \to 0$.

$= \displaystyle \large \lim_{\frac{h}{2} \,\to\, 0} \normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \cos{\Bigg[\dfrac{2x+h}{2}\Bigg]}$

Find the value of the function when $x$ tends to $0$. According to $\lim sinx/x$ as $x$ approaches $0$ rule, the value of the first multiplying factor is one and substitute $h = 0$ in the second multiplying factor.Remember, the entire expression in limit form represents the derivative of $\sin{x}$ with respect to $x$ formula.

$\implies$ $\dfrac{d}{dx} \, \sin{x} \,=\, 1 \times \cos{\Bigg[\dfrac{2x+0}{2}\Bigg]}$

$\implies$ $\dfrac{d}{dx} \, \sin{x} \,=\, \cos{\Bigg[\dfrac{2x}{2}\Bigg]}$

$\implies$ $\require{cancel} \dfrac{d}{dx} \, \sin{x} \,=\, \cos{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx} \, \sin{x} \,=\, \cos{x}$

It is proved that the derivative (or) differentiation of $\sin{x}$ with respect to $x$ is $\cos{x}$.