Chain rule formula
There is a formula for using the chain rule, when y is a function of u and u is a function of x:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
The formula can also be written in function notation,
if \(y = f(g(x))\) then\(\frac{dy}{dx} = f'(g(x))g'(x)\)
Examples using the formula and function notation
Let's look at some examples of the chain rule to help you understand it further:
If \(y = (2x - 1)^3\) find \(\frac{dy}{dx}\)
First, you can start by looking at the formula for the chain rule before rewriting your y in terms of both y and u:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
\(y = (u)^3\) \(u = 2x -1\)
Next you can take your y and u and differentiate them to find: \(\frac{dy}{du} \space \frac{du}{dx}\)
\(y = (u)^3\)
\(\frac{dy}{du} = 3u^2\)
Now you can differentiate your u to find : \(\frac{du}{dx}\)
\(u = 2x - 1\)
\(\frac{du}{dx} = 2\)
Now that you have each aspect of the formula you can find \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
\(\frac{dy}{dx} = 3u^2 \cdot 2\)
\(\frac{dy}{dx} = 6u^2\)
Lastly, you need to make sure your answer is written in terms of x, to do this you can substitute in \(u = 2x-1\):\(\frac{dy}{dx} = 6u^2\)
\(\frac{dy}{dx} = 6(2x -1)^2\)
The question may also involve some trigonometric functions. Let's look at an example of how to work through it.
If \(y = (\sin x)^5\) find \(\frac{dy}{dx}\)
You can start this just like before, finding each aspect of your formula:
\(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\)
\(y = (u)^5\) \(u = \sin x\)
Next you can differentiate both y and u to find \(\frac{dy}{du}\) and \(\frac{du}{dx}\):
\(\frac{dy}{du} = 5u^4\) \(\frac{du}{dx} = \cos x\)
Now that you have all the aspects you can solve to find : \(\frac{dy}{dx}\)
\[\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\]
\[\frac{dy}{dx} = 5u^4 \cdot \cos x\]
Once again, you need to make sure your answer is written in terms of x. To do this, you have to substitute back in \(u = \sin x\):\[\frac{dy}{dx} = 5u^4 \cdot \cos x\]\[\frac{dy}{dx} = 5(\sin x)^4 \cdot \cos x\]
You may be given the question in function notation form and asked to differentiate.
Differentiate \(f(g(x)) = (3x^2 + 2)^2\)
First, you need to start by looking at your function notation formula:
If \(y = f(g(x))\) then\(\frac{dy}{dx} = f'(g(x))g'(x)\)
Now you can identify your f(x) and g(x):\(f(x) = x^2\) \(g(x) = 3x^2 + 2\)
Next, you can differentiate f(x) and g(x) to find f'(x) and g'(x):
\(f'(x) = 2x \qquad g'(x) = 6x\)
For the formula you also need to find: f'(g(x))
\(f'(g(x)) = 2(3x^2 + 2)\)
Now that you have every aspect of the function notation formula, you can substitute each part in and find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = f'(g(x))g'(x)\)
\(\begin{align} \frac{dy}{dx} &=2(3x^2 + 2)(6x) \\ &= (6x^2 + 4)(6x) \\ &= 36x^3 + 24x \end{align}\)
What if the function is not in the form y = f(x)
It is important to consider the formula you would use if the function you are given is not in the form \(y = f(x)\). The formula to use for this is:
\(\frac{dy}{dx} = \frac{1}{dx/dy}\)
The question could look something like this:
Find the value of \(\frac{dy}{dx}\) at the point (4, 1) on the curve \(y^4 + 2y = x\).
Let's work through this question to see how you would solve it. First, you can start by differentiating the equation with respect to y:
\(y^4 + 2y = x\)
\(\frac{dx}{dy} = 4y^3 +2\)
Next, you substitute your differentiated equation into the formula,
\[\frac{dy}{dx} = \frac{1}{dx/dy}\]
\(\frac{dy}{dx} = \frac{1}{4y^3 + 2}\)
Now all you need to do is substitute the y from the point on the curve from the question into the formula to find your answer:
\[\frac{dy}{dx} = \frac{1}{4y^3 + 2}\]
\[\frac{dy}{dx} = \frac{1}{4(1)^3 + 2}\]
\[\frac{dy}{dx} = \frac{1}{6}\]
Find the value of \(\frac{dy}{dx}\) at the point (6, 3) on the curve \(4y^2 + 3y = x\)
Once again you start by differentiating the equation with respect to y:
\(4y^2 + 3y = x\)
\(\frac{dx}{dy} = 8y + 3\)
Now you can input that into the formula to find the value of \(\frac{dy}{dx}\) at the point (6,3): \[\frac{dy}{dx} = \frac{1}{dx/dy}\]
\[\frac{dy}{dx} = \frac{1}{8y+3}\]
Next you substitute the y value from the coordinates in order to solve the equation:
\[\frac{dy}{dx} = \frac{1}{8y+3}\]
\[\frac{dy}{dx} = \frac{1}{8(3)+3}\]
\[\frac{dy}{dx} = \frac{1}{27}\]
What is the reverse chain rule?
The reverse chain rule is used when integrating a function; it involves taking the differentiated function and taking it back to its original form.
Integrate \(\int{12(3x+3)^3 dx}\)
To do this, you can start by identifying your main function and breaking it down to revert it to its original integral. You can do this by working backwards:
\(12(3x + 3)^3\)
\(4(3x + 3)^3 \cdot 3\)
\((3x + 3)^4\)
\(\int{12(3x + 3)^3 dx} = (3x + 3)^4 + c\)
Above is a breakdown of how to get to the answer. When differentiating x to a power, you can bring down the power in front of x and the power decreases by 1, for example x3 becomes 3x2. You also know that something has been multiplied together to get the 12 – in this instance, 4 since the power is 3. Taking it one step further back, you can take the 4 back up to a power. When using the reverse chain rule, it is also important that you add a constant to your answer, represented by c.
Chain Rule - Key takeaways
The chain rule is a rule used for differentiating composite functions, and these functions are also known as a function of a function.
The formula that can be used when differentiating using the chain rule is:
\(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\).
The formula can also be written in function notation, if \(y = f'(g(x))\) then \(\frac{dy}{dx} = f'(g(x))g'(x)\)
The chain rule can also be used if the composite function involves trigonometric functions.
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Frequently Asked Questions about Chain Rule
What is the chain rule?
The chain rule is a rule used in differentiating functions.
When do you use the chain rule?
The chain rule can be used when differentiating a composite function, also known as a function of a function.
What is the reverse chain rule?
The reverse chain rule is used when integrating a function, it involves taking a differentiated function back to its integral.
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