# 5.1 Differentiation (first principles, rules) and sketching graphs

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## Differentiation from first principles

The tangent problem has given rise to the branch of calculus called differential calculus and the equation: ${lim}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$ defines the derivative of the function $f\left(x\right)$ . Using [link] to calculate the derivative is called finding the derivative from first principles .

Derivative

The derivative of a function $f\left(x\right)$ is written as ${f}^{\text{'}}\left(x\right)$ and is defined by:

${f}^{\text{'}}\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

There are a few different notations used to refer to derivatives. If we use the traditional notation $y=f\left(x\right)$ to indicate that the dependent variable is $y$ and the independent variable is $x$ , then some common alternative notations for the derivative are as follows: ${f}^{\text{'}}\left(x\right)={y}^{\text{'}}=\frac{dy}{dx}=\frac{df}{dx}=\frac{d}{dx}f\left(x\right)=Df\left(x\right)={D}_{x}f\left(x\right)$

The symbols $D$ and $\frac{d}{dx}$ are called differential operators because they indicate the operation of differentiation , which is the process of calculating a derivative. It is very important that you learn to identify these different ways of denoting the derivative, and that you are consistent in your usage of them when answering questions.

Though we choose to use a fractional form of representation, $\frac{dy}{dx}$ is a limit and is not a fraction, i.e. $\frac{dy}{dx}$ does not mean $dy÷dx$ . $\frac{dy}{dx}$ means $y$ differentiated with respect to $x$ . Thus, $\frac{dp}{dx}$ means $p$ differentiated with respect to $x$ . The ` $\frac{d}{dx}$ ' is the “operator", operating on some function of $x$ .

Calculate the derivative of $g\left(x\right)=x-1$ from first principles.

1. We know that the gradient at a point $x$ is given by: ${g}^{\text{'}}\left(x\right)={lim}_{h\to 0}\frac{g\left(x+h\right)-g\left(x\right)}{h}$

2. $g\left(x+h\right)=x+h-1$

3. $\begin{array}{ccc}\hfill {g}^{\text{'}}\left(x\right)& =& \underset{h\to 0}{lim}\frac{g\left(x+h\right)-g\left(x\right)}{h}\hfill \\ & =& \underset{h\to 0}{lim}\frac{x+h-1-\left(x-1\right)}{h}\hfill \\ & =& \underset{h\to 0}{lim}\frac{x+h-1-x+1}{h}\hfill \\ & =& \underset{h\to 0}{lim}\frac{h}{h}\hfill \\ & =& \underset{h\to 0}{lim}1\hfill \\ & =& 1\hfill \end{array}$
4. The derivative ${g}^{\text{'}}\left(x\right)$ of $g\left(x\right)=x-1$ is 1.

## Derivatives

1. Given $g\left(x\right)=-{x}^{2}$
1. determine $\frac{g\left(x+h\right)-g\left(x\right)}{h}$
2. hence, determine ${lim}_{h\to 0}\frac{g\left(x+h\right)-g\left(x\right)}{h}$
3. explain the meaning of your answer in (b).
2. Find the derivative of $f\left(x\right)=-2{x}^{2}+3x$ using first principles.
3. Determine the derivative of $f\left(x\right)=\frac{1}{x-2}$ using first principles.
4. Determine ${f}^{\text{'}}\left(3\right)$ from first principles if $f\left(x\right)=-5{x}^{2}$ .
5. If $h\left(x\right)=4{x}^{2}-4x$ , determine ${h}^{\text{'}}\left(x\right)$ using first principles.

## Rules of differentiation

Calculating the derivative of a function from first principles is very long, and it is easy to make mistakes. Fortunately, there are rules which make calculating the derivative simple.

## Investigation : rules of differentiation

From first principles, determine the derivatives of the following:

1. $f\left(x\right)=b$
2. $f\left(x\right)=x$
3. $f\left(x\right)={x}^{2}$
4. $f\left(x\right)={x}^{3}$
5. $f\left(x\right)=1/x$

You should have found the following:

 $f\left(x\right)$ ${f}^{\text{'}}\left(x\right)$ $b$ 0 $x$ 1 ${x}^{2}$ $2x$ ${x}^{3}$ $3{x}^{2}$ $1/x={x}^{-1}$ $-{x}^{-2}$

If we examine these results we see that there is a pattern, which can be summarised by: $\frac{d}{dx}\phantom{\rule{4pt}{0ex}}\left({x}^{n}\right)=n{x}^{n-1}$

There are two other rules which make differentiation simpler. For any two functions $f\left(x\right)$ and $g\left(x\right)$ : $\frac{d}{dx}\left[f\left(x\right)±g\left(x\right)\right]={f}^{\text{'}}\left(x\right)±{g}^{\text{'}}\left(x\right)$ This means that we differentiate each term separately.

The final rule applies to a function $f\left(x\right)$ that is multiplied by a constant $k$ . $\frac{d}{dx}\left[k.f\left(x\right)\right]=k{f}^{\text{'}}\left(x\right)$

Determine the derivative of $x-1$ using the rules of differentiation.

1. We will apply two rules of differentiation: $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$ and $\frac{d}{dx}\left[f\left(x\right)-g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right)\right]-\frac{d}{dx}\left[g\left(x\right)\right]$

2. In our case $f\left(x\right)=x$ and $g\left(x\right)=1$ . ${f}^{\text{'}}\left(x\right)=1$ and ${g}^{\text{'}}\left(x\right)=0$

3. The derivative of $x-1$ is 1 which is the same result as was obtained earlier, from first principles.

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