1.7.1 Basic Differentiation
In this topic we will learn how to:
- use the notationsf'(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 + 1Before 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} = 2Now 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 2Which 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 + 7Let’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 + 7Let’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 + 7Let’s differentiate the third term,
2(3)x^{2 - 1}6xLet’s move on to the fourth term,
y = 5x^{4} + 2x^{3} + 3x^{2} + \textcolor{#2192ff}{x} + 7Let’s differentiate the fourth term,
(1)x^{1 - 1}1The 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 + 12. 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} + 5Let’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}}