# Proof of Constant multiple rule of Derivatives

Let $f(x)$ be a function in a variable $x$. In differential calculus, the differentiation of the function $f(x)$ with respect to $x$ is written in the following mathematical form.

$\dfrac{d}{dx}{\, f(x)}$

For deriving the derivative of a constant multiple function with respect to a variable, we must know the fundamental definition of the differentiation of a function in limit form.

### Definition of the derivative in Limit form

According to the definition of the derivative, the derivative of a function $f(x)$ with respect to $x$ can be written in limit form.

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

If we take the change in variable $x$ is equal to $h$, which means $\Delta x = h$, then the equation can be expressed in terms of $h$ as follows.

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

### Define a Constant multiple function

Let $k$ be a constant, then the product of constant $k$ and the function $f(x)$ is called the constant multiple function, which is written in product form as $k.f(x)$ mathematically.

### Find the Derivative of constant multiple function

Take, the constant multiple function is denoted by $g(x)$. Therefore, $g(x) = k.f(x)$. Now, write the differentiation of $g(x)$ with respect to $x$ in limit form as per the definition of the derivative.

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

In this case, $g(x) = k.f(x)$ then $g(x+h) = k.f(x+h)$. Now, substitute them in the above equation.

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

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

$k$ is a common factor in the expression of the numerator of the function and it can be taken out as a common factor from the terms as per the factorization by taking out the common factors.

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

Now, factorize the function for separating the constant.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[k \times \Bigg(\dfrac{f(x+h)-f(x)}{h}\Bigg) \Bigg]}$

It can be further simplified by the product rule of limits.

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

Now, evaluate the limit of the constant by the direct substitution method.

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

According to the fundamental definition of the derivative, the limit of the second function is the differentiation of the function $f(x)$.

$=\,\,\,$ $k \times \dfrac{d}{dx}{\, f(x)}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \Big(k.f(x)\Big)}$ $\,=\,$ $k.\dfrac{d}{dx}{\, f(x)}$

Therefore, it is proved that the derivative of a constant multiple function with respect to a variable is equal to the product of the constant and the derivative of the function.

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.