Math Doubts

Derivative of cosx Proof

The derivative of cos function with respect to a variable is equal to negative sine. If $x$ denotes a variable, then the cosine function is expressed as $\cos{x}$ in mathematics. So, the differentiation of the $\cos{x}$ with respect to $x$ is equal to $-\sin{x}$ and it is proved from first principle mathematically.

Beginner’s Method

On the basis of definition of the derivative, the derivative of the function with respect to $x$ can be written in the following limiting operation form.

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

Take $f{(x)} = \cos{x}$, then $f{(x+h)} = \cos{(x+h)}$. Now, the mathematical proof for the differentiation of $\cos{x}$ function with respect to $x$ can be derived from first principle.

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

Simplify the Trigonometric expression

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

The simplification of the trigonometric expression in the numerator can be started by expanding the $\cos{(x+h)}$ function as per the angle sum identity of cos function.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}\cos{h}-\sin{x}\sin{h}-\cos{x}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}\cos{h}-\cos{x}-\sin{x}\sin{h}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}{(\cos{h}-1)}-\sin{x}\sin{h}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}{((-1)(1-\cos{h}))}-\sin{x}\sin{h}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{(-\cos{x})}{(1-\cos{h})}-\sin{x}\sin{h}}{h}}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{{(-\cos{x})}{(1-\cos{h})}-\sin{x}\sin{h}}{h}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{{(-\cos{x})}{(1-\cos{h})}}{h} – \dfrac{\sin{x}\sin{h}}{h}\Bigg]}$

According to difference rule of limits, the limit of the difference of two functions can be evaluated by evaluating the difference of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{(-\cos{x})}{(1-\cos{h})}}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{x}\sin{h}}{h}}$

The input of limiting operations is in terms of $h$. So, the functions in terms of $x$ are constant functions. Therefore, $-\cos{x}$ from first function and $\sin{x}$ from second function can be separated according to the constant multiple rule of limits.

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $-$ ${(\sin{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{h}}{h}}$

Evaluate Limit of Trigonometric function

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $-$ ${(\sin{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{h}}{h}}$

According to the limit of trigonometric function rule, the limit of $\sin{h}/{h}$ as $h$ approaches $0$ is equal to one.

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $-$ ${(\sin{x})} \times 1$

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $-\sin{x}$

Continue simplifying the expression

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1-\cos{h}}{h}}$ $-\sin{x}$

The trigonometric expression in the numerator in the first term of the expression can be simplified by the power reduction rule of sine function.

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{h}{2}\Big)}}{h}}$ $-\sin{x}$

The number $2$ is a factor and multiplying sine function in the numerator and it divides the denominator. It is recommendable to shift it to denominator for simplifying the term further.

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}}$ $-\sin{x}$

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{h}{2}\Big)} \times \sin{\Big(\dfrac{h}{2}\Big)}}{\dfrac{h}{2}}}$ $-\sin{x}$

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

The limit of product of two functions can be written as product of their limits by the product rule of limits.

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

Evaluate Limits of trigonometric functions

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

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. So, $\dfrac{h}{2} \to 0$. Therefore, if $h$ approaches $0$, then $\dfrac{h}{2}$ also approaches $0$.

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

Take $t = \dfrac{h}{2}$ and transform the first limit function only in the first term of the expression.

$= \,\,\,$ ${(-\cos{x})} \times \Bigg[ \displaystyle \large \lim_{t \,\to\, 0}{\normalsize \dfrac{\sin{t}}{t}} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)} \Bigg]}$ $-\sin{x}$

Therefore, the limit of the quotient of $\sin{t}$ by $t$ as $t$ tends to $0$ is equal to one.

$= \,\,\,$ ${(-\cos{x})} \times \Bigg[1 \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)} \Bigg]}$ $-\sin{x}$

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

$= \,\,\,$ ${(-\cos{x})} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \sin{\Big(\dfrac{h}{2}\Big)}}$ $-\sin{x}$

Now, evaluate the limit of the trigonometric function by the direct substitution method.

$= \,\,\,$ ${(-\cos{x})} \times \sin{\Big(\dfrac{0}{2}\Big)}$ $-\sin{x}$

$= \,\,\,$ ${(-\cos{x})} \times \sin{(0)}$ $-\sin{x}$

The exact value of the sine of zero degrees is zero.

$= \,\,\,$ ${(-\cos{x})} \times 0$ $-\sin{x}$

$= \,\,\,$ $0-\sin{x}$

$= \,\,\,$ $-\sin{x}$

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

Therefore, it is proved by first principle that the derivative of $\cos{x}$ with respect to $x$ is equal to negative $\sin{x}$.

Advanced Method

According to definition of the derivative, the differentiation of the function in terms of $x$ can be written in the following limiting operation form.

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

Take $f{(x)} = \cos{x}$, then $f{(x+\Delta x)} = \cos{(x+\Delta x)}$. Now, the proof of derivative of $\cos{x}$ function with respect to $x$ can be started by the first principle.

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

Try difference to product conversion rule

Now, use difference to product identity of cos functions to combine the difference of two cosine functions in the numerator of the function.

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

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

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

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{-1 \times 2 \times \sin{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\Delta x}}$

Simplify the entire function

The factor $2$ multiplies the other factors in the numerator and it divides the denominator. So, it can be shifted to denominator.

$= \,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{-1 \times \sin{\Bigg[\dfrac{2x+\Delta x}{2}\Bigg]}\sin{\Bigg[\dfrac{\Delta x}{2}\Bigg]}}{\dfrac{\Delta x}{2}}}$

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

Now, divide the whole function as product of two functions.

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

As per the product rule of limits, the limit of product of functions is equal to product of their limits.

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

Find Limit of the function

Firstly, evaluate the limit of the sin function by using direct substitution method.

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

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

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

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

The limit of second sine function is almost similar to the limit of sinx/x as x approaches 0 rule but its input is slightly different. Therefore, set this function same as the standard result.

If, $\Delta x \,\to\, 0$, then $\dfrac{\Delta x}{2} \,\to\, \dfrac{0}{2}$. So, $\dfrac{\Delta x}{2} \,\to\, 0$. It proved that if $\Delta x$ approaches $0$, then $\dfrac{\Delta x}{2}$ also tends to zero.

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

Take $m = \dfrac{\Delta x}{2}$ and then transform the entire function in terms of $m$.

$= \,\,\,$ $-\sin{x}$ $\times$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin{m}}{m}}$

The limit of trigonometric function is same as the limit of $\sin{x}/x$ as $x$ tends to $0$ formula exactly. Therefore, the limit of trigonometric function is equal to $1$.

$= \,\,\,$ $-\sin{x} \times 1$

$= \,\,\,$ $-\sin{x}$

$\therefore \,\,\,\,\,\, {(\cos{x})}’ \,=\, -\sin{x}$

Therefore, it has proved from first principle that the differentiation of cos function with respect to a variable is equal to negative sine.



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