Quotient Rule

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

To begin you can look at the formula and find each part that you need:

\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

Next, you can substitute each variable that you have found into the formula:

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

You're now able to simplify and solve your formula to find :

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

If \(y = \frac{\sin x}{3x + 5}\) find \(\frac{dy}{dx}\).

As before, it is good to start by identifying the formula you would need and breaking it down to find each part of the equation. You know that because there is a fraction involved in the question you can use the quotient rule formula. Let's take a look at the formula and find each part of it:

\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

\(u = \sin x\) \(v = 3x +5\) \(\frac{du}{dx} = \cos x\) \(\frac{dv}{dx} = 3\)

Now that you have identified each part of the formula you can substitute the parts into the equation to solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{(3x +5)(\cos x) - (\sin x)(3)}{(3x + 5)^2}\)

If \(f(x) = \frac{x}{3x^3 + 2}\) find \(f'(x)\).

Once again it is good to start by identifying the formula needed and each part of it. Since there is a quotient involved and the question is written in function form, you know that you need to use the quotient rule in function notation:

\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)

\(g(x) = x\) \(h(x) = 3x^3 + 2\) \(g'(x) = 1\) \(h'(x) = 9x^2\)

Next, you can substitute each part you have identified into the formula to find \(f'(x)\):

\(f'(x) = \frac{(3x^3 + 2)(1) - (x)(9x^2)}{(3x^3 + 2)^2}\)

\(f'(x) = \frac{(3x^3 + 2) - (9x^3)}{(3x^3 + 2)^2}\)

\(f'(x) = \frac{-6x^3 + 2}{(3x^3 + 2)^2}\)

If \(f(x) = \frac{2x + 2}{\ln x}\) find \(f'(x)\).

You can start by looking at your formula for the function notation of the quotient rule and getting each part of the equation ready to solve:

\(f(x) = \frac{g(x)}{h(x)}\) then\(f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\)

\(g(x) = 2x +2\) \(h(x) = \ln x\) \(g'(x) = 2\) \(h'(x) = \frac{1}{x}\)

Now you can substitute each part into the formula to solve for \(f'(x)\):

\(f'(x) = \frac{(\ln x) (2) - (2x + 2) (\frac{1}{x})}{(\ln x)^2}\) \(f'(x) = \frac{2x \ln x - 2x -2}{x(\ln x)^2}\)

Find the value of \(\frac{dy}{dx}\) for the point (2, 1/3) on the curve where \(y = \frac{x^2}{3x+6}\)

For this type of question you would still start in the same way as before, identify your formula and find each part of it:

\(\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

\(u = x^2\) \(v = 3x +6\) \(\frac{du}{du} = 2x\) \(\frac{dv}{dx} = 3\)

Again, just like before you substitute each part into the formula to solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{(3x + 6) (2x) - (x^2)(3)}{(3x +6)^2}\)

\(\frac{dy}{dx} = \frac{6x^2 + 12x - 3x^2}{(3x +6)^2}\)

\(\frac{dy}{dx} = \frac{3x^2 + 12x}{(3x +6)^2}\)

Now, because you are looking to find the value of \(\frac{dy}{dx}\) when the point of the curve is (2, 1/3), you can substitute the x coordinate into the equation above:

\(\frac{dy}{dx} = \frac{3x^2 + 12x}{(3x +6)^2}\)

\(\frac{dy}{dx} = \frac{3(2)^2 + 12(2)}{(3(2) +6)^2}\)

\(\frac{dy}{dx} = \frac{36}{144}\)