Product Rule

If \(y = 5xe^2\) find \(\frac{dy}{dx}\):

First, you can start by looking at your product rule formula and finding each aspect of the formula:

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

If \(y = 5xe^2\), \(u = 5x\) and \(v = e^2\)

To find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\) you can differentiate your u and v:

\(\frac{du}{dx} = 5\) \(\frac{dv}{dx} = 0\)

\[\frac{dy}{dx} = (5x)(0) + (e^2)(5) = 5e^2\]

If \(y = (4\sin{x})e^2\) find \(\frac{dy}{dx}\)

As before, you can start by finding each part of the formula:

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

If \(y = (4\sin{x})e^2\), \(u = 4\sin x\) and \(v = e^2\)

Now you can differentiate your u and v to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):

\(\frac{du}{dx} = 4\cos x\) \(\frac{dv}{dx} = 0\)

Finally, you can substitute each part into your formula to find \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = (4\sin x)(0) + (e^2)(4\cos x)\]

\[\frac{dy}{dx} = 4e^2\cos x\]

If \(f(x) = 2x^2(x^2 + 4)\) find \(f'(x)\)

Once again, you can start by breaking down the formula for the product rule in function notation and finding each part.

\(f'(x) = g(x)h'(x) + h(x)g'(x)\)

If \(f(x) = 2x^2(x^2 + 4)\), \(g(x) = 2x^2\) and \(h(x) = x^2 + 4\)

Next, you can differentiate g(x) and h(x) to find the derivatives, g'(x) and h'(x):

\(g' = 4x\) \(h'(x) = 2x\)

Now that you have each part of the formula, you can solve it to find f'(x):

\(f'(x) = (2x^2)(2x) + (x^2 + 4)(4x) = 2(4x^3 + 8x)\)

If \(y = \ln x(x^2)\) find \(\frac{dy}{dx}\)

To begin with, let's look at the product rule formula and find each aspect of it:

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

If \(y = \ln x(x^2)\), \(u = \ln x\) and \(v = x^2\)

Let's differentiate the terms to find \(\frac{dv}{dx}\) and \(\frac{du}{dx}\):

\(\frac{du}{dx} = \frac{1}{x}\) \(\frac{dv}{dx} = 2x\)

Now you can put each part into the formula to find \(\frac{dy}{dx}\):

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

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