Math Doubts

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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