Quotient Rule

The quotient rule is a rule used when you are differentiating a quotient function. A quotient function can be described as a function that is being divided by another function.

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An example of a quotient function is $$y = \frac{3x}{2x + 2}$$ or $$y = \frac{x^2}{3x}$$.

Quotient rule formula

There is a formula that can be used when using the quotient rule to differentiate It is shown below:

If $$y = \frac{u}{v}$$ then$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

This formula can also be written in function notation:

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

Examples using the quotient rule formula

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:

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

$$u = 2x^2$$ $$v = 2x +2$$ $$\frac{du}{dx} = 4x$$ $$\frac{dv}{dx} = 2$$

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 :

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

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

Let's look at an example where a trigonometric function is involved to see how you would solve for $$\frac{dy}{dx}$$:

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 $$\frac{dy}{dx}$$:

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

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

Examples using the function notation

It is useful to also know how to use the quotient rule in terms of its function notation as this may be how it appears within the exam question. Let's remind ourselves of the formula for the function notation before working through some examples!

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

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 $$f'(x)$$:

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

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

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

Let's work through another example.

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 $$f'(x)$$:

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

How to solve problems using the quotient rule

Since functions can be represented visually using graphs, sometimes you may need to solve a question based on the points the function may cross. To do this, you can still simply use the quotient rule formula if it applies, then with some extra steps afterwards you will be able to find the value.

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:

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

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

$$\frac{dy}{dx} = \frac{36}{144}$$

Quotient Rule - Key takeaways

• The quotient rule is a rule used in differentiation. It is used when you are differentiating a quotient, which is a function that is being divided by another function.

• The formula for the quotient rule is $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$ if $$y = \frac{u}{v}$$

• The formula can also be written in function notation when $$f(x) = \frac{g(x)}{h(x)}$$ then$$f(x) = \frac{g(x)}{h(x)}$$ then$$f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}$$

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What is the quotient rule?

The quotient rule is a rule that is used when differentiating.

When do you use the quotient rule?

You can use the quotient rule when you are differentiating a quotient function, this is when a function is being divided by another function.

What is the formula for the quotient rule?

The formula for the quotient rule is dy/dx = (v du/dx - u dv/dx)/v^2 if y = u/v  , this can also be written in function notation, when f(x)=g(x)/h(x) then f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2

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