Product Rule & Quotient Rule
What is the product rule?
The product rule is a formula which will help us find the derivative of the product of two functions. Suppose we have two functions, f(x) and g(x) which are being multiplied together such that: f(x)g(x).
What would the derivative of this product be? This is where the product rule will help us, and it is defined as:
Product Rule:
[latex]\large{\frac{d}{dx}f(x)g(x) = f(x)g'(x)+f'(x)g(x)}[/latex]
Note that f'(x) and g'(x) are the derivatives of f(x) and g(x) respectively. The above formula will enable us to compute the derivatives of functions which consist of the product of two other functions. Let us now use the product rule to solve the following example question:
If [latex]h(x) = x^2ln(x)[/latex], find [latex]h'(x)[/latex].
First, we need to identify that the given function is a product of two other functions. We know that the product rule can help us find the derivative of two functions being multiplied together. In this case, we have:
[latex]h(x) = f(x)g(x)[/latex]
Then, the product rule tells us that:
[latex]h'(x) = f(x)g'(x) + f'(x)g(x)[/latex].
From the question, it is quite easy to see that [latex]f(x) = x^2[/latex] and [latex]g(x) = ln(x)[/latex]. So, to use the product rule, we also need to determine f'(x) and g'(x), which are [latex]f'(x) = 2x[/latex] and [latex]g'(x) = \frac{1}{x}[/latex].
Now, all we need to do is substitute these into the product rule equation, as shown below:
[latex]h'(x) = f(x)g'(x) +f'(x)g(x)[/latex]
[latex]h'(x) = (x^2)(\frac{1}{x})+(2x)(ln(x))[/latex]
[latex]h'(x) = x+2xln(x)[/latex]
This shows how easy it is to use the derivative product rule to find the derivative of a function that is a product of two other functions!
Example
Find the derivative of the following function:
[latex]\large{h(x) = (x-3x^2)cos(x)}[/latex]
Step 1:
The first step is to identify that two functions are being multiplied together, which can be labeled as:
[latex]\large{f(x) = x-3x^2}[/latex] and [latex]\large{g(x) =cos(x) }[/latex]
To use the product rule, we also need the derivative of these two functions, which should be easy for us to do by now:
[latex]\large{f'(x) = 1 – 6x}[/latex] and [latex]\large{g'(x) = -sin(x)}[/latex]
Step 2:
Now, applying the product rule, we have:
[latex]\large{h'(x) = f(x)g'(x)+f'(x)g(x) = (x-3x^2)(-sin(x)) + (1-6x)(cos(x)) }[/latex]
Simplifying, we have:
[latex]\large{ h'(x) = [-xsin(x)+3x^2sin(x)] + [cos(x)-6xcos(x)]}[/latex]
Then, after simplifying again:
[latex]\large{h'(x) = 3x^2sin(x) -xsin(x)-6xcos(x)+cos(x) }[/latex]
Notice how we can factor out sin(x) from the first two terms and cos(x) from the second two terms. Doing this, give us our final answer for the derivative:
[latex]\large{h'(x) = sin(x)(3x^2-x)+cos(x)(1-6x)}[/latex]
Step 1:
The first step is to identify that two functions are being multiplied together, which can be labeled as:
[latex]\large{f(x) = x-3x^2}[/latex] and [latex]\large{g(x) =cos(x) }[/latex]
To use the product rule, we also need the derivative of these two functions, which should be easy for us to do by now:
[latex]\large{f'(x) = 1 – 6x}[/latex]
and
[latex]\large{g'(x) = -sin(x)}[/latex]
Step 2:
Now, applying the product rule, we have:
[latex]h'(x) = f(x)g'(x)+f'(x)g(x)[/latex]
[latex]\small{ = (x-3x^2)(-sin(x)) + (1-6x)(cos(x))}[/latex]
Simplifying, we have:
[latex] \scriptsize{h'(x) = [-xsin(x)+3x^2sin(x)] + [cos(x)-6xcos(x)]}[/latex]
Then, after simplifying again:
[latex]\scriptsize{h'(x) = 3x^2sin(x) -xsin(x)-6xcos(x)+cos(x)}[/latex]
Notice how we can factor out sin(x) from the first two terms and cos(x) from the second two terms. Doing this, give us our final answer for the derivative:
[latex]\small{h'(x) = sin(x)(3x^2-x)+cos(x)(1-6x)}[/latex]
What is the Quotient Rule?
The quotient rule is a formula which will help us find the derivative of the quotient of two functions. Suppose we have two functions, f(x) and g(x) which are being divided together as such: [latex]\frac{f(x)}{g(x)}[/latex].
The quotient rule will help us find the derivative of this, which is defined as:
Quotient Rule:
[latex]\Large{\frac{d}{dx}(\frac{f(x)}{g(x)}) = \frac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}}[/latex]
Just like how the product rule helps us find the derivative of the product of two functions, the quotient rule will help us find the derivative of the quotient of two functions.
Let us now use the quotient rule to solve the following question:
If [latex]\large{h(x) = \frac{6x^4}{5x^2+2}}[/latex], find [latex]\large{h'(x)}[/latex].
First, we can see that the given function h(x) is a quotient of two other functions. We can label these functions as
[latex]\large{f(x) = 6x^4}[/latex] and [latex]\large{g(x) = 5x^2+2}[/latex].
Similar to the product rule, to use the quotient rule formula, we also need to determine f'(x) and g'(x). This should be easy to do for us at this point, and we have:
[latex]\large{f'(x) = 24x^3}[/latex] and [latex]\large{g'(x) = 10x}[/latex].
Now, all we need to do is substitute these into the quotient rule equation as shown below:
[latex]\large{h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}}[/latex]
[latex]\large{h'(x) = \frac{(5x^2+2)(24x^3)+(6x^4)(10x)}{(5x^2+2)^2}}[/latex]
[latex]\large{h'(x) = \frac{120x^5+48x^3-60x^5}{25x^4+20x^2+4}}[/latex]
[latex]\large{h'(x) = \frac{60x^5+48x^3}{25x^4+20x^2+4}}[/latex]
That is how you use the quotient rule to find the derivative of a function that is the quotient of two other functions.
Example
Find the derivative of the following function:
[latex]h(x) = \Large{\frac{sin(x)+x^3}{e^x-2x^2}}[/latex]
Step 1:
The first step is to identify that two functions are being divided, which can be labeled as:
[latex]\large{f(x) = sin(x)+x^3}[/latex] and [latex]\large{g(x) =e^x-2x^2 }[/latex]
To use the quotient rule, we also need the derivative of these two functions, which should be simple:
[latex]\large{f'(x) = cos(x)+3x^2}[/latex] and [latex]\large{g'(x) = e^x-4x}[/latex]
Step 2:
Now, applying the quotient rule, we have:
[latex]\large{h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2} = \frac{(e^x-2x^2)(cos(x)+3x^2) \:- \:(sin(x)+x^3)(e^x-4x)} {(e^x-2x^2)^2} }[/latex]
Simplifying, we have:
[latex]\Large{ h'(x) = \frac{e^xcos(x)+e^x3x^2-2x^2cos(x)-6x^4-e^xsin(x)+4xsin(x)-e^xx^3+4x^4}{(e^x-2x^2)^2}}[/latex]
Finally, after simplifying fully, we end up with:
[latex]\Large{h'(x) = \frac{cos(x)+3x^2}{e^x-2x^2} \;-\; \frac{(e^x-4x)(sin(x)+x^3)}{(e^x-2x^2)^2} }[/latex]
Step 1:
The first step is to identify that two functions are being divided, which can be labeled as:
[latex]\large{f(x) = sin(x)+x^3}[/latex]
and
[latex]\large{g(x) =e^x-2x^2 }[/latex]
To use the quotient rule, we also need the derivative of these two functions, which should be simple:
[latex]\large{f'(x) = cos(x)+3x^2}[/latex]
and
[latex]\large{g'(x) = e^x-4x}[/latex]
Step 2:
Now, applying the quotient rule, we have:
[latex]\large{h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}}[/latex]
[latex]\large{ = \frac{(e^x-2x^2)(cos(x)+3x^2) \:- \:(sin(x)+x^3)(e^x-4x)} {(e^x-2x^2)^2}}[/latex]
Simplifying, we have:
[latex] h'(x) = [/latex]
[latex]\small{\frac{e^xcos(x)+e^x3x^2-2x^2cos(x)-6x^4-e^xsin(x)+4xsin(x)-e^xx^3+4x^4}{(e^x-2x^2)^2}}[/latex]
Finally, after simplifying fully, we end up with:
[latex]\small{h'(x) = \frac{cos(x)+3x^2}{e^x-2x^2} \;-\; \frac{(e^x-4x)(sin(x)+x^3)}{(e^x-2x^2)^2} }[/latex]
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