Differentiation Rules

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 \(u = 2x^3 + x\):

\[\frac{dy}{dx} = 6(2x^3 + x)^5(6x^2 +1)\]

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 \(u = \sin x\):

\[\frac{dy}{dx} = 3u^2 \cdot \cos x\]

\[\frac{dy}{dx} = 3(\sin x)^2 \cos x\]

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

\[\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}\]$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$

\[\frac{dy}{dx} = (x^2)(\cos x) + (\sin x)(2x)\]

\[\frac{dy}{dx} = \cos x(x^2) + 2x\sin x\]

\[\frac{dy}{dx} = x^2\cos x + 2x \sin x\]

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:

\(\frac{dy}{dx} = \frac{(3x +4) \cdot 2 - (2x) \cdot 3}{(3x + 4)^2}\)

\(\frac{dy}{dx} = \frac{8}{(3x + 4)^2}\)

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

\[\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}\]

\[\frac{dy}{dx} = \frac{(2x^2)(-\sin x) - (\cos x)(4x)}{(2x^2)^2}\]

\[\frac{dy}{dx} = \frac{-2x^2 \sin x - 4x \cos x}{(2x^2)^2}\]