In this section we define the derivative of a function.
Given a function \(f(x)\), its derivative is another function whose output value at any value of \(x\) equals the gradient of the curve \(y=f(x)\) at that same value of \(x\).
This is the "starting point" to our studies of calculus and more particularly of differentiation.
We start by learning what the derivative function is as well as gain some insight in what we'll be able to achieve with derivatives.
We then learn the formal definition of the derivative function \(f'(x)\) and see how to differentiate from first principles.
In the following tutorial we learn about the derivative function. In particular we learn what it is and what it is used for.
Although we don't learn how to find the derivative function in this tutorial it gives us a good understanding of what we'll be using derivatives and differentiation for. Make sure to watch it.
Given a function \(f(x)\), we define its derivative function as a second function whose value at any value of \(x\), say \(x=c\), equals to the gradient of the curve \(y=f(x)\) at the point along its curve at which \(x=c\).
Consequently, the derivative of a function (we usually just say "the derivative") is often referred to as its gradient function.
The derivative can be referred to using either of the following two notations:
\[f'(x)\quad \text{or} \quad \frac{dy}{dx}\]
Say we need to calculate the gradient of a curve \(y=f(x)\) at the point along its length with \(x\)-coordinate \(x=2\), we would calculate \(f'(2)\) and that would equal to the gradient of the curve at that point.
The process via which we find the derivative function is known as differentiation. We'll often see phrases such as "find the derivative", "differentiate the following function", or even
"find the gradient function". Given a function \(f(x)\), all those phrases refer to the same task: finding the derivative function \(f'(x)\).
The function defined by \(f(x)=\frac{x^3}{3}+x-2\) has derivative \(f'(x)=x^2+1\).
Using what you know so far about the derivate function, find the gradient of the curve \(y=\frac{x^3}{3}+x-2\) at:
To find the gradient of the curve at any value of \(x\), we use the derivative function \(f'(x)\).
The formula for differentiating functions from first principles, relies on the following fact:
The gradient of a curve at any point along its length equals to the gradient of the tangent to the curve at that same point.
Given a function \(f(x)\), its derivative function \(f'(x)\) can be obtained using the following: \[f'(x)=\lim_{h \to 0 } \begin{bmatrix} \frac{f\begin{pmatrix}x+h \end{pmatrix}-f \begin{pmatrix} x \end{pmatrix}}{h} \end{bmatrix}\] Using the fact that the derivative function can also be written \(\frac{dy}{dx}\), we'll sometimes see this definition of the derivative function written: \[\frac{dy}{dx}=\lim_{h \to 0 } \begin{bmatrix} \frac{f\begin{pmatrix}x+h \end{pmatrix}-f \begin{pmatrix} x \end{pmatrix}}{h} \end{bmatrix}\]
Differentiate the function defined by \(f(x)=x^2\) from first principles.
Given the function \(f(x)=x^3\).
Given the function defined by the curve: \[y=3x^2+1\]
The derivative function is: \(f'(x)=2x\)
Working from first principles, we write:
\[f'(x) = \lim_{h \to 0 } \begin{bmatrix} \frac{f\begin{pmatrix}x+h \end{pmatrix}-f \begin{pmatrix} x \end{pmatrix}}{h} \end{bmatrix}\]
Since \(f(x)=x^2\), \(f\begin{pmatrix}x+h\end{pmatrix}=\begin{pmatrix}x+h\end{pmatrix}^2\) and the definition leads us to:
\[f'(x) = \lim_{h \to 0 } \begin{bmatrix} \frac{\begin{pmatrix}x+h \end{pmatrix}^2-x^2}{h} \end{bmatrix}\]
We now simplify as much as possible:
\[\begin{aligned} f'(x) & = \lim_{h \to 0 } \begin{bmatrix} \frac{\begin{pmatrix}x+h \end{pmatrix}^2-x^2}{h} \end{bmatrix} \\
& = \lim_{h \to 0 } \begin{bmatrix} \frac{x^2+2xh+h^2-x^2}{h} \end{bmatrix} \\
& = \lim_{h \to 0 } \begin{bmatrix} \frac{2xh+h^2}{h} \end{bmatrix} \\
f'(x) & = \lim_{h \to 0 } \begin{bmatrix} 2x+h \end{bmatrix} \end{aligned}\]
The final step is to "take the limit".
As \(h\) tends towards \(0\), \(2x+h\) will tend towards \(2x\) and so, in the limit:
\[f'(x) = \lim_{h \to 0 } \begin{bmatrix} 2x+h \end{bmatrix} =2x\]
So the derivative of \(f(x)=x^2\) is \(f'(x)=2x\).