# Proof of $\dfrac{d}{dx}{\,\sin{(ax)}}$ by the Definition of the Derivatives

The derivative of sine of a multiple angle $a$ times $x$ with respect to $x$ is equal to the product of multiple constant $a$ and cosine of angle $a$ times $x$.

$\dfrac{d}{dx}\,\sin{(ax)}$ $\,=\,$ $a\cos{(ax)}$

The derivative of the sine of a multiple angle is used as a formula in differential calculus and it can be derived from the fundamental definition of the derivatives.

### Use the Definition of the Derivatives

According to the first principle of the differentiation, the derivative of the sine of angle $a$ times $x$ with respect to $x$ is written in the following limiting operations in mathematics.

$(1).\,\,\,$ $\dfrac{d}{dx}\,\sin{(ax)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x\,\to\,0}{\normalsize \dfrac{\sin{\big(a(x+\Delta x)\big)}-\sin{(ax)}}{\Delta x}}$

$(2).\,\,\,$ $\dfrac{d}{dx}\,\sin{(ax)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\big(a(x+h)\big)}-\sin{(ax)}}{h}}$

### Simplify the Rational function

In the first term of the numerator, the binomial $x+h$ is multiplied by its coefficient $a$. According to the distributive property of multiplication over the addition, the factor $a$ can be distributed across the sum of the terms $x$ and $h$.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{(ax+ah)}-\sin{(ax)}}{h}}$

The trigonometric expression in the numerator is an expression that expresses the difference of the two sine functions. According to the difference to product identity in sine functions, the difference of sine functions can be calculated by transforming it into the product form of trigonometric functions.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{ax+ah+ax}{2}\bigg)}\sin{\bigg(\dfrac{ax+ah-ax}{2}\bigg)}}{h}}$

Now, let’s focus on simplifying the trigonometric expression in the numerator of the rational function by the sum and difference rules of expressions.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{ax+ax+ah}{2}\bigg)}\sin{\bigg(\dfrac{ax-ax+ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{2ax+ah}{2}\bigg)}\sin{\bigg(\dfrac{\cancel{ax}-\cancel{ax}+ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{2ax+ah}{2}\bigg)}\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

### Evaluate the Limit of the function

The rational function is successfully simplified and it is time to calculate the limit of the simplified mathematical function as the value of $h$ is closer to $0$.

The direct substitution is not recommendable method at this time because it gives us an indeterminate form. So, let us split the rational function as a product of two functions by splitting the function as a product of two functions.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{2\cos{\bigg(\dfrac{2ax+ah}{2}\bigg)} \times \sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

There is no limit rule in cosine function but there is a trigonometric limit rule in sine function. So, split the rational function as a product of two functions as follows.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(2\cos{\bigg(\dfrac{2ax+ah}{2}\bigg)} \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}\Bigg)}$

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

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize 2\cos{\bigg(\dfrac{2ax+ah}{2}\bigg)}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

Now, let us focus on finding the limits of the functions. Firstly, let’s try to find the limit of the first function. Use the direct substitution method to find the limit of the function as the value of $h$ approaches to zero.

$=\,\,\,$ $2\cos{\bigg(\dfrac{2ax+a(0)}{2}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{2ax+a \times 0}{2}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{2ax+0}{2}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{2ax}{2}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{2 \times ax}{2}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{\bigg(\dfrac{\cancel{2} \times ax}{\cancel{2}}\bigg)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}}$

Now, it is right time to concentrate on finding the limit of the remaining function. The function is in terms of sine function. So, it is better to convert into the trigonometric limit rule in sine function.

The expression in the denominator should be same as the angle inside the sine function for applying the trigonometric limit rule in sine function to the rational function. Hence, let’s make an adjustment to convert the rational function into the required form.

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(1 \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{\Big(\dfrac{a}{2}\Big)}{\Big(\dfrac{a}{2}\Big)} \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{\Big(\dfrac{a}{2}\Big) \times 1}{\Big(\dfrac{a}{2}\Big)} \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\Big(\dfrac{a}{2}\Big) \times \dfrac{1}{\Big(\dfrac{a}{2}\Big)} \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{h}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\Big(\dfrac{a}{2}\Big) \times \dfrac{1 \times \sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{a}{2}\Big) \times h}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\Big(\dfrac{a}{2}\Big) \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{a \times h}{2}\Big)}\Bigg)}$

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\Big(\dfrac{a}{2}\Big) \times \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}\Bigg)}$

The factor $a$ divided by $2$ is a constant and it can be taken out from the limit operation as per the constant multiple rule of limits.

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\Big(\dfrac{a}{2}\Big)$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

Now, let’s multiply the first two factors in order to simplify the trigonometric expression further.

$=\,\,\,$ $2\cos{(ax)}$ $\times$ $\dfrac{a}{2}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $\dfrac{2\cos{(ax)} \times a}{2}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $\dfrac{2 \times \cos{(ax)} \times a}{2}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $\dfrac{\cancel{2} \times \cos{(ax)} \times a}{\cancel{2}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $\cos{(ax)} \times a$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $a \times \cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

$=\,\,\,$ $a\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

The function in sine function with limiting operation is almost similar to the trigonometric limit rule. It indicates that the trigonometric limit rule in sine function can be applied but the input value for the limit operation should be same as either denominator or the angle inside the sine function.

$(1).\,\,\,$ If $h\,\to\,0$, then $a \times h\,\to\,a \times 0$. Therefore, $ah\,\to\,0$

$(2).\,\,\,$ If $ah\,\to\,0$, then $\dfrac{ah}{2}\,\to\,\dfrac{0}{2}$. Therefore $\dfrac{ah}{2}\,\to\,0$

The two steps have cleared that if $h$ is closer to zero, then $a$ times $h$ divided by $2$ also approaches to $0$.

$=\,\,\,$ $a\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{\Large \frac{ah}{2} \large \,\to\,0}{\normalsize \dfrac{\sin{\bigg(\dfrac{ah}{2}\bigg)}}{\Big(\dfrac{ah}{2}\Big)}}$

Let’s denote $a$ times $h$ divided by $2$ by a literal $y$ for our convenience.

$=\,\,\,$ $a\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{(y)}}{(y)}}$

$=\,\,\,$ $a\cos{(ax)}$ $\times$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{y}}{y}}$

According to the trigonometric limit rule in sine function, the limit of the sine of angle $y$ divided by $y$ as the value of $y$ tends to $0$ is equal to one.

$=\,\,\,$ $a\cos{(ax)}$ $\times$ $1$

$=\,\,\,$ $a\cos{(ax)}$

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Jun 26, 2023

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