# Derivative of cosx formula Proof

The differentiation of cosx function can be derived mathematically in calculus by expressing derivative of cosx function in limit form.

### Express Differentiation of function in Limit form

The differentiation of any function can be expressed by writing the differential expression in limit form according to principle method.

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

Take $f(x) = \cos{x}$ then $f(x+h) = \cos{(x+h)}$

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

### Use sum to product transformation rule

Now, use sum to product transformation formula to simplify the numerator of the expression.

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

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

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

### An adjustment for further simplification

The angle of the second sine function is $\dfrac{h}{2}$. It contains $h$ as denominator and $2$ as multiplying factor in numerator. They can be expressed as a ratio of them to apply lim sinx/x as x approaches 0 rule in upcoming step.

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

Now, split the expression as two multiplying factors.

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

### Apply Limit Product Rule

The limit $h$ tends to zero condition belongs to both multiplying factors. So, it can be applied to both multiplying factors.

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

The second multiplying factor is almost same as the lim sinx/x as x approaches 0 formula. In order to apply this rule, the limit of the second multiplying function should be adjusted appropriately.

If $h \to 0$ then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. Hence, $h \to 0$ can be replaced by $\dfrac{h}{2} \to 0$ in the second multiplying factor.

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

### Simplifying the mathematical expression

Substitute $h = 0$ in the first multiplying factor and apply limit sinx/x as x approaches 0 rule for the second multiplying factor. The limit expression actually represents the derivative of cosx with respect to x formula.

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

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

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

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

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

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