1.7.1 Basic Differentiation

In this topic we will learn how to:

• use the notations$$f'(x),\ \ f''(x),\ \ \frac{dy}{dx}\ \ \textmd{and}\ \ \frac{d^{2}y}{dx^{2}}$$
• use the derivative of $x^{n}$ (for any rational $n$), together with constant multiples, sums and differences of functions, and of composite functions using the chain rule

Differentiation is a form of calculus in which we determine the derivative of any function. The derivative of any function is its gradient, it is denoted by,
$$\frac{dy}{dx}$$If the function is denoted by $f(x)$ then $\frac{dy}{dx}$ is written as,
$$f'(x)$$Once you find the first derivative of a function, you have also found the gradient function.
$$\textbf{\textcolor{gray}{The Second Derivative}}$$If you differentiate the first derivative, you get the second derivative. This is denoted by,
$$\frac{d^{2}y}{dx^{2}}$$For the function $f(x)$, it would be written as,
$$f''(x)$$The second derivative is used to determine the nature of stationary points i.e whether a stationary point is a maximum or minimum turning point.

$$\textcolor{gray}{\textbf{Differentiation of }ax^{n}}$$To differentiate $y = ax^{n}$, where $a$ is a non-zero constant we use the general formula,
$$\frac{dy}{dx} = anx^{n - 1}$$Note: If you differentiate a constant, you get zero.
$$\textcolor{gray}{\textbf{Differentiation of }(ax + b)^{n}}$$To differentiate $y = (ax + b)^{n}$ we use the general formula,
$$\frac{dy}{dx} = an(ax + b)^{n - 1}$$This is the formula we get from the chain rule, however, for more complex functions it may be wiser to use the chain rule.
$$\textbf{\textcolor{gray}{The Chain Rule}}$$The chain rule gives us a way to calculate the derivative of a composition of functions.
Let’s look at an example.

Example 1
Find the derivative of $y = (2x + 1)^{3}$
$$y = (\textcolor{#2192ff}{2x + 1})^{3}$$First identify the inner function i.e whatever is inside the bracket, and equate it to $u$,
$$u = \textcolor{#2192ff}{2x + 1}$$Substitute $u$ into the original equation,
$$y = (\textcolor{#2192ff}{2x + 1})^{3}$$$$y = \textcolor{#2192ff}{u}^{3}$$This is the outer function.

We now have two separate equations, which we can differentiate easily,
$$y = u^{3}\ \ \ \ \ \ u = 2x + 1$$Before we differentiate, we can define a chain rule. A chain rule, is an equation that allows to find our required differential, by considering the derivatives of the inner and outer functions separately. To define a chain rule, start by writing out the derivative you want to find,
$$\frac{dy}{dx}$$Equate it to the product of the derivatives of the inner and outer functions,
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$Note: If you were to simplify the right-hand side, you would get $\frac{dy}{dx}$.

Now let’s differentiate the two equations to evaluate $\frac{dy}{du}$ and $\frac{du}{dx}$,
$$y = u^{3}\ \ \ \ \ \ u = 2x + 1$$$$\frac{dy}{du} = 3u^{2}\ \ \ \ \ \ \frac{du}{dx} = 2$$Now we can use the chain rule, we defined earlier,
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$Substitute into the chain rule,
$$\frac{dy}{dx} = 3u^{2} \times 2$$Which simplifies to give,
$$\frac{dy}{dx} = 6\textcolor{#2192ff}{u}^{2}$$Replace $u$,
$$\frac{dy}{dx} = 6\textcolor{#2192ff}{(2x + 1)}^{2}$$Your final differential would be,
$$\frac{dy}{dx} = 6(2x + 1)^{2}$$

Let’s walk through some examples to cement the concepts we have learnt above.

1. Differentiate $y = 5x^{4} + 2x^{3} + 3x^{2} + x + 7$.
$$y = \textcolor{#2192ff}{5x^{4}} + 2x^{3} + 3x^{2} + x + 7$$Let’s start by differentiating the first term,
$$4(5)x^{4 - 1}$$$$20x^{3}$$Let’s move on to the second term,
$$y = 5x^{4} + \textcolor{#2192ff}{2x^{3}} + 3x^{2} + x + 7$$Let’s differentiate the second term,
$$3(2)x^{3 - 1}$$$$6x^{2}$$Let’s move on to the third term,
$$y = 5x^{4} + 2x^{3} + \textcolor{#2192ff}{3x^{2}} + x + 7$$Let’s differentiate the third term,
$$2(3)x^{2 - 1}$$$$6x$$Let’s move on to the fourth term,
$$y = 5x^{4} + 2x^{3} + 3x^{2} + \textcolor{#2192ff}{x} + 7$$Let’s differentiate the fourth term,
$$(1)x^{1 - 1}$$$$1$$The fifth term, $7$, gives a $0$ when it is differentiated. Therefore, putting everything together, the final answer would be,
$$\frac{dy}{dx} = 20x^{3} + 6x^{2} + 6x + 1$$2. It is given that the function $f$ is defined by $f(x) = \frac{4}{x^{\frac{1}{2}}} - \frac{2}{3x^{\frac{3}{2}}}$. Find $f'(x)$.
$$f(x) = \frac{4}{\textcolor{#2192ff}{x^{\frac{1}{2}}}} - \frac{2}{3\textcolor{#2192ff}{x^{\frac{3}{2}}}}$$The first step, is to use the laws of indices to bring up the parts containing $x$,
$$f(x) = 4\textcolor{#2192ff}{x^{-\frac{1}{2}}} - \frac{2}{3}\textcolor{#2192ff}{x^{-\frac{3}{2}}}$$Note: We cannot differentiate if any part containing $x$ is in the denominator.

Now we can differentiate,
$$f(x) = \textcolor{#2192ff}{4x^{-\frac{1}{2}}} - \frac{2}{3}\textcolor{#2192ff}{x^{-\frac{3}{2}}}$$Let’s start with the first term,
$$-\frac{1}{2}(4)x^{-\frac{1}{2} - 1}$$$$-2x^{-\frac{3}{2}}$$Let’s move on to the second term,
$$f(x) = 4x^{-\frac{1}{2}}\ \textcolor{#2192ff}{-\ \frac{2}{3}x^{-\frac{3}{2}}}$$Note: The negative sign is part of the second term.

Let’s differentiate the second term,
$$-\left(-\frac{3}{2}\right)\left(\frac{2}{3}\right)x^{-\frac{3}{2} - 1}$$$$x^{-\frac{5}{2}}$$Putting everything together,
$$f'(x) = -2x^{-\frac{3}{2}} + 3x^{-\frac{5}{2}}$$Note: It is adviseable to make the powers in $x$ positive, however, it is not mandatory.

Therefore, the final answer would be,
$$f'(x) = -\frac{2}{x^{\frac{3}{2}}} + \frac{1}{x^{\frac{5}{2}}}$$3. It is given that $y = \sqrt{2x^{3} + 5}$. Find $\frac{dy}{dx}$.
$$y = \sqrt{2x^{3} + 5}$$Let’s use laws of indices to rewrite the square root sign as the power $\frac{1}{2}$,
$$y = (\textcolor{#2192ff}{2x^{3} + 5})^{\frac{1}{2}}$$To differentiate this function we need to use the chain rule. Let’s identify the inner function and equate it to $u$,
$$u = \textcolor{#2192ff}{2x^{3} + 5}$$Let’s write the outer function in terms of $u$,
$$y = (\textcolor{#2192ff}{2x^{3} + 5})^{\frac{1}{2}}$$$$y = u^{\frac{1}{2}}$$We now have two functions that are easy to differentiate,
$$y = u^{\frac{1}{2}}\ \ \ \ \ \ u = 2x^{3} + 5$$Let’s use the derivatives of the two functions to define a chain rule for $\frac{dy}{dx}$,
$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$Let’s differentiate our two functions to evaluate $\frac{dy}{du}$ and $\frac{du}{dx}$,
$$y = u^{\frac{1}{2}}\ \ \ \ \ \ u = 2x^{3} + 5$$$$\frac{dy}{du} = \frac{1}{2}u^{\frac{1}{2} - 1}\ \ \ \ \ \ \frac{du}{dx} = 3(2)x^{3 - 1}$$$$\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}\ \ \ \ \ \ \frac{du}{dx} = 6x^{2}$$Let’s go back to the chain rule we defined earlier,
$$\frac{dy}{dx} = \textcolor{#2192ff}{\frac{dy}{du}} \times \textcolor{#0f0}{\frac{du}{dx}}$$Substitute in the expressions of $\frac{dy}{du}$ and $\frac{du}{dx}$,
$$\frac{dy}{dx} = \textcolor{#2192ff}{\frac{1}{2}u^{-\frac{1}{2}}} \times \textcolor{#0f0}{6x^{2}}$$Simplify,
$$\frac{dy}{dx} = 3x^{2}u^{-\frac{1}{2}}$$Substitute $u$,
$$\frac{dy}{dx} = 3x^{2}(2x^{3} + 5)^{-\frac{1}{2}}$$Therefore, the final answer is,
$$\frac{dy}{dx} = 3x^{2}(2x^{3} + 5)^{-\frac{1}{2}}$$