Math Doubts

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 $\cos{x}$.

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