There are many different rules that can be used when differentiating, and each rule can be used for a specific reason. There are three different rules that you will need to know:
It is important to note that not all of the below formulae are written in the provided formula booklet and you will have to memorise them for your exam.
Chain rule
The chain rule can be used when you are differentiating a composite function, which is also known as a function of a function. The formula for this rule is below. This is when y is a function of u and u is a function of x:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
Using the chain rule, differentiate \(y = (2x^3 + x)^6\)
First, you can start by rewriting it in terms of y and u:
\[\begin{align}y = (u)^6 \\ u = 2x^3 + x \end{align}\]
Now you can find the first part of your chain rule formula \(\frac{dy}{du}\), by differentiating your y:
\(\frac{dy}{du} = 6 u^5\)
Next, you can find the second part of your chain rule formula \(\frac{du}{dx}\), by differentiating your u:
\[\frac{du}{dx} = 6x^2 + 1\]
Now that you have found both parts of your sum you can multiply them together to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
\[\frac{dy}{dx} = 6u^5 \cdot 6x^2 + 1\]
\[\frac{dy}{dx} = 6u^5 (6x^2 + 1)\]
Finally, it is important to express your answer in terms of x, and to do this you can use \(u = 2x^3 + x\):
\[\frac{dy}{dx} = 6(2x^3 + x)^5(6x^2 +1)\]
The function that you are given may involve a trigonometric function. Let's work through an example to see how this would be solved.
If \(y = (\sin x)^3\) find \(\frac{dy}{dx}\)
Let’s start by identifying your u and y:
\(y = (u)^3\) \(u = \sin x\)
Now you can look at the chain rule formula, break it down, and find each part:
\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\(\frac{dy}{du} = 3u^2\) \(\frac{du}{dx} = \cos x\)
Next, you can put your \(\frac{dy}{du}\) and \(\frac{du}{dx}\) into the formula to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
\[\frac{dy}{dx} = 3u^2 \cdot \cos x\]
Finally, you need to make sure your answer is written in terms of x, and you can do this by substituting in \(u = \sin x\):
\[\frac{dy}{dx} = 3u^2 \cdot \cos x\]
\[\frac{dy}{dx} = 3(\sin x)^2 \cos x\]
The chain rule can also be written in notation form, which allows you to differentiate a function of a function:
If \(y = f(g(x))\) then \(\frac{dy}{dx} = f'(g(x)) g'(x)\)
Product rule
The product rule is used when you are differentiating the product of two functions. A product of a function can be defined as two functions being multiplied together. When using this rule you need to make sure you have the product of two functions and not a function of a function, as they can be confused. The formula for this rule is below – this is if y=uv when u and v are functions of x:
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\)
This function can also be written in function notation:
If \(f(x) = g(x)h(x)\) then \(f'(x) = g(x)h'(x) + h(x)g'(x)\)
Given that \(y = 2xe^2\) find \(\frac{dy}{dx}\)
Looking at the formula, first, you need to identify each part of the formula:
\[\frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
\(u = 2x\) \(v = e^2\)
To find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\) you need to differentiate your u and v:
\(\frac{du}{dx} = 2\) \(\frac{dv}{dx} = 0\)
Now you have found each part of the formula you can solve for \(\frac{dy}{dx}\):
\begin{align} \frac{dy}{dx} = (2x)(0) + (e^2)(2) \\ \frac{dy}{dx} = 2e^2 \end{align}
Given that \(y = x^2 \sin x\) find \(\frac{dy}{dx}\)
To do this you can start by looking at the formula and identifying what you need:
\[\frac{dy}{dx} = u \frac{dv}{dx} + v\frac{du}{dx}\]
\(u = x^2\) \(v = \sin x\)
Now you can differentiate your u and v to find the next part of the formula:
\(\frac{du}{dx} = 2x\) \(\frac{dv}{dx} = \cos{x}\)
Input all of the information into the formula to find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]
\[\frac{dy}{dx} = (x^2)(\cos x) + (\sin x)(2x)\]
\[\frac{dy}{dx} = \cos x(x^2) + 2x\sin x\]
\[\frac{dy}{dx} = x^2\cos x + 2x \sin x\]
Quotient rule
The quotient rule is used when you are differentiating the quotient of two functions, this is when one function is being divided by another. The formula used for this rule is below, this is when \(y = \frac{u}{v}\):
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)
This can also be written in function notation:
If \(f(x) = \frac{g(x)}{h(x)}\) then \(f'(x) = \frac {h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)
If \(y = \frac{2x}{3x +4}\) find \(\frac{dy}{dx}\) :
First, you can start by looking at your formula and find each part of it:
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)
\(u = 2x\) \(v = 3x + 4\) \(\frac{du}{dx} = 2\) \(\frac{dv}{dx} = 3\)
Now you can solve the formula with all of the above information:
\(\frac{dy}{dx} = \frac{(3x +4) \cdot 2 - (2x) \cdot 3}{(3x + 4)^2}\)
\(\frac{dy}{dx} = \frac{8}{(3x + 4)^2}\)
Let’s look at a second example involving a trigonometric function.
If \(y = \frac{\cos x}{2x^2}\) find \(\frac{dy}{dx}\).
Just like in the previous example, you can start by looking at your formula and finding each part of it to find your \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)
\(u = \cos x\) \(v = 2x^2\) \(\frac{du}{dx} = -\sin x\) \(\frac{dy}{dx} = 4x\)
Now that you have each part of the formula you can substitute them into the formula and find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}\]
\[\frac{dy}{dx} = \frac{(2x^2)(-\sin x) - (\cos x)(4x)}{(2x^2)^2}\]
\[\frac{dy}{dx} = \frac{-2x^2 \sin x - 4x \cos x}{(2x^2)^2}\]
Differentiation Rules - Key takeaways
There are three main differentiation rules, chain rule, product rule, and quotient rule.
Each rule is used for a different reason and has a different formula for you to use.
The chain rule is used when you are differentiating a composite function.
The product rule is used when you are differentiating the products of two functions.
The quotient rule is used when you are differentiating the quotient of two functions.
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