# Derivative of sin x with respect to x

## Formula

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

It is read as the derivative of sine of angle $x$ with respect to $x$ is equal to cosine of angle $x$.

### Proof

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

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

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

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

Use sum to product identity of sine to express them in product form.

$$\implies \dfrac{d}{dx} \, \sin x = \lim_{h \to 0} \dfrac{2 \sin \Bigg[\dfrac{x+h -x}{2}\Bigg] \cos \Bigg[\dfrac{x+h – (-x)}{2}\Bigg]}{h}$$

$$\require{cancel} \implies \dfrac{d}{dx} \, \sin x = \lim_{h \to 0} \dfrac{2 \sin \Bigg[\dfrac{\cancel{x}+h -\cancel{x}}{2}\Bigg] \cos \Bigg[\dfrac{x+h+x}{2}\Bigg]}{h}$$

$$\implies \dfrac{d}{dx} \, \sin x = \lim_{h \to 0} \dfrac{2 \sin \Bigg[\dfrac{h}{2}\Bigg] \cos \Bigg[\dfrac{x+h+x}{2}\Bigg]}{h}$$

$$\implies \dfrac{d}{dx} \, \sin x = \lim_{h \to 0} \dfrac{2 \sin \Bigg[\dfrac{h}{2}\Bigg] \cos \Bigg[\dfrac{2x+h}{2}\Bigg]}{h}$$

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

$$\implies \dfrac{d}{dx} \, \sin x = \lim_{h \to 0} \dfrac{\sin \Bigg[\dfrac{h}{2}\Bigg] \cos \Bigg[\dfrac{2x+h}{2}\Bigg]}{\dfrac{h}{2}}$$

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

Use product law of limits to apply the limit to both functions.

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

The sine function consists $\dfrac{h}{2}$ as angle and the denominator also has it. 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, write $h \to 0$ by $\dfrac{h}{2} \to 0$.

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

The ratio of sine of angle to angle when the value of angle approaches zero is equal to one.

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

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

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

$\implies \dfrac{d}{dx} \, \sin x = 1 \times \cos x$

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

It is proved that the derivative of $\sin x$ with respect to $x$ is $\cos x$.