## 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 *x ^{3}* becomes

*3x*

^{2}. 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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