Differentiation 1

Differentiation is a method for calculating the gradient of a curve at a given point. On this sheet, we recap the definition of the derivative of a function in one variable and practice the method of differentiation from first principles.

We will also revise the standard formula for the derivative of a power, and practice using it to differentiate polynomials. The problem set at the end of this resource includes some contextual questions to give a taste of how differentiation can be used to solve problems in the real world.

First Principles Differentiation

Definition 1.

The derivative of a function $f$ at a point $P=(a,f(a))$ on the curve $y=f(x)$ is

$$f^{\prime }(a)=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}.$$ (1)

The rationale for this formula is the following: suppose $Q$ is a second point on the curve, very close to $P$.

Figure 1: Two nearby points on a curve, connected by a straight line.

Provided $P$ and $Q$ are very close together, the line $L$ passing through both of these points will give a good approximation for the tangent line at $P$.

Suppose that $P=(a,f(a))$ and $Q=(a+h,f(a+h))$, where $h$ is a very small number.

The gradient of the straight line through $P$ and $Q$ is given by the standard formula

$$\text{Gradient of }L=\frac{\text{Change in }y}{\text{Change in }x}=\frac{f(a+h)-f(a)}{h}$$

As $h$ becomes very small, the point $Q$ approaches $P$, and $L$ becomes the tangent line at $P$. The gradient of $L$ becomes the derivative of $f$ at $P$, given by formula (1).

Differentiation using (1) is sometimes called differentiation by first principles.

Since we want to view the derivative as a funciton, we tend to use $x$ instead of $a$ in formula (1). Then, the derivative is given by

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$ (2)

Example 2.

Differentiate $f(x)=3x^2$ from first principles.

Solution.

We use (2). Before we can evaluate the limit, we calculate $\frac{f(x+h)-f(x)}{h}$ for this particular function $f(x)=3x^2$.

We have

${\displaystyle \frac{f(x+h)-f(x)}{h}}$ $={\displaystyle \frac{3{(x+h)}^2-3x^2}{h}}$

$={\displaystyle \frac{3(x^2+2xh+h^2)-3x^2}{h}}$

$={\displaystyle \frac{3x^2+6xh+3h^2-3x^2}{h}}$

$={\displaystyle \frac{6xh+3h^2}{h}}=6x+3h.$

Letting $h$ tend to zero, we then get

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}(6x+h)=6x.$$

Exercise 3.

Let $f(x)=x^3$.

1. Calculate ${\displaystyle \frac{f(x+h)-f(x)}{h}}$.

2. Hence use (2) to differentiate $f(x)$ from first principles.

Some Properties of Derivatives

Linearity Property

If two functions $f(x)$ and $g(x)$ are differentiable, and $a$ and $b$ are constants, then the derivative of $af(x)+bg(x)$ is $$\frac{d}{dx}\left(af(x)+bg(x)\right)=a{f}^{\prime }(x)+b{g}^{\prime }(x).$$

Differentiating Powers

General rule: For any non-zero real number $r$, $$\frac{d}{dx}\left(x^r\right)=r{x}^{r-1}.$$ (3)

Equation (3) may be used in conjunction with the linearity property for derivatives to differentiate any linear combination of powers of $x^r$.

Example 4.

Use equation (3) to differentiate $f(x)=3x^2-2\sqrt{x}+\frac{7}{x^3}$.

Solution.

First we rewrite the expression for $f(x)$ using power notation:

$$f(x)=3x^2-3x^{\frac{1}{2}}+7x^{-3}.$$

By the linearity property for derivatives, we can calculate $f^{\prime }(x)$ by differentiating term by term. Each term can be differentiating using the power rule (3). Hence

$f^{\prime }(x)$ $=6x-3\cdot {\displaystyle \frac{1}{2}}{x}^{-\frac{1}{2}}+7(-3)x^{-4}$

$=6x-{\displaystyle \frac{3}{2\sqrt{x}}}-{\displaystyle \frac{21}{x^4}}.$

Exercise 5.

Use (3) to differentiate $15x-{\displaystyle \frac{2}{\sqrt[3]{x}}}-{\displaystyle \frac{x^4}{2}}$ with respect to $x$.