# Why is derivative of constant zero?

Dec 22, 2015

The derivative represents the change of a function at any given time.

Take and graph the constant $4$:
graph{0x+4 [-9.67, 10.33, -2.4, 7.6]}

The constant never changes—it is constant.

Thus, the derivative will always be $0$.

Consider the function ${x}^{2} - 3$.
graph{x^2-3 [-9.46, 10.54, -5.12, 4.88]}

It is the same as the function ${x}^{2}$ except that it's been shifted down $3$ units.
graph{x^2 [-9.46, 10.54, -5.12, 4.88]}

The functions increase at exactly the same rate, just in a slightly different location.

Thus, their derivatives are the same—both $2 x$. When finding the derivative of ${x}^{2} - 3$, the $- 3$ can be disregarded since it does not change the way in which the function changes.

Dec 22, 2015

Use the power rule: $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$

A constant, say $4$, can be written as

$4 {x}^{0}$

Thus, according to the power rule, the derivative of $4 {x}^{0}$ is

$0 \cdot 4 {x}^{-} 1$

which equals

$0$

Since any constant can be written in terms of ${x}^{0}$, finding its derivative will always involve multiplication by $0$, resulting in a derivative of $0$.

Dec 22, 2015

Use the limit definition of the derivative:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

If $f \left(x\right) = \text{C}$, where $\text{C}$ is any constant, then

$f \left(x + h\right) = \text{C}$

Thus,

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\text{C"-"C}}{h} = {\lim}_{h \rightarrow 0} \frac{0}{h} = {\lim}_{h \rightarrow 0} 0 = 0$