Chain Rule

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 \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]

\(\frac{dy}{dx} = 3u^2 \cdot 2\)

\(\frac{dy}{dx} = 6u^2\)

\(\frac{dy}{dx} = 6u^2\)

\(\frac{dy}{dx} = 6(2x -1)^2\)

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 : \(\frac{dy}{dx}\)

\[\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\]

\[\frac{dy}{dx} = 5u^4 \cdot \cos x\]

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\(\frac{dy}{dx} = f'(g(x))g'(x)\)

\(f(x) = x^2\) \(g(x) = 3x^2 + 2\)

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 \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = f'(g(x))g'(x)\)

\(\begin{align} \frac{dy}{dx} &=2(3x^2 + 2)(6x) \\ &= (6x^2 + 4)(6x) \\ &= 36x^3 + 24x \end{align}\)

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:

\[\frac{dy}{dx} = \frac{1}{8y+3}\]

\[\frac{dy}{dx} = \frac{1}{8(3)+3}\]

\[\frac{dy}{dx} = \frac{1}{27}\]

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.