# First Principles at a Given $x$-value

## First Principles at a Given x-Value

Suppose we are given a function $f(x)$ and asked to find its derivative at the point where $x=a$. This is actually the gradient of the tangent to the curve at $x=a$, which we write as $f'(a)$.
There are two methods for finding $f'(a)$ using first principles:

### Method 1

The first method is to start with the definition of the gradient function.
$$f'(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$$

### Example 1

Use the first principles formula $\displaystyle f'(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$ to find the instantaneous rate of change in $f(x)=x^2-2x$ at the point where $x=3$.

### Method 2

The second method to consider two points on the graph of $y=f(x)$, a fixed point $\color{teal}A(a,f(a))$ and a variable point $\color{teal}B(x,f(x))$. The gradient of chord $\color{teal}A\color{teal}B=\dfrac{f(x)-f(a)}{x-a}$.
In the limit as $\color{teal}B$ approaches $\color{teal}A$, $x \rightarrow a$ and the gradient of chord $\color{teal}A\color{teal}B$ approaches gradient of the tangent at $\color{teal}A$. $$f'(a)=\lim_{x \rightarrow a}\dfrac{f(x)-f(a)}{x-a}$$ This is an alternative definition of the gradient of the tangent at $x=a$. Note that the gradient of the tangent at $x=a$ is defined as the gradient of the curve at the point where $x=a$, and is the instantaneous rate of change in $y$ with respect to $x$ at that point.

### Example 2

Use the first principles formula $\displaystyle f'(a)=\lim_{x \rightarrow a}\dfrac{f(x)-f(a)}{x-a}$ to find the instantaneous rate of change in $f(x)=2x^2+9$ at the point where $x=2$.