The Product Rule

By:Mohammad-Ali Bandzar | Dec 25 2019

The product rule is a formal rule for differentiating problems where one function is multiplied by another. The rule follows from the limit definition of derivative and states that the derivative of loading... will be loading.... This rule only works if f and g are both differentiable functions. This pattern works for any number of functions as i will demonstrate below.

With A Single Function

with just one function say loading... we can arbitarly define a second function loading... and multiply the two together to create a new function we will call loading...can you see how loading... is the same function as loading...? Now if we take the derivative of loading... we get loading... now if we recall our constant rule, we know that the derivative of a constant in this case 1 will be zero. So if we sub in zero we get loading... now if we sub in loading... we get loading... we have now shown that we can use our product rule for derivatives on a single function by defining a second function equal to one.

With Three Functions

Let loading... assuming that f,g,h are differentiable functions. we can arbitarily define a new function loading... and we can rewrite loading... as loading... we can now use our two function definition of the product rule to find the derivative, loading... we can now independantly find the derivitive of loading... using the same method loading.... we can now plug loading... back into our derivative of loading... to get: loading... this pattern is infinitely long and if you are wondering why you can read about it in the proof below.

Proof

To start off our proof we will write out the definition of the derivative:

loading...

loading...

we can now add and subtract loading... from the numerator which is allowed becuase by adding then subtracting the same thing we will not change our end result.

loading...

we will then factor loading... from the first and second last terms

loading...

we will then factor loading... from our remaining 2 terms in the numerator

loading...

Now if we recall the addition property of limits:

if a is a real number and both loading... and loading... exist then loading...

since we have already stated that loading... and loading... must be differentiable functions we can apply this rule here to expand our equation:

loading...

Now if we recall the multiplicative property of limits:

As long as both loading... and loading... exist then loading...

since we have already stated that loading... and loading... must be differentiable functions we can apply this rule here to expand our equation:

loading...

We can now simplify the left most limit, since loading... is differentiable and since all differentiable functions are continuous we can conclude that loading... must be continuous allowing us to sub what h is approaching directly into loading... giving us loading.... We can now substitute that into our equation:

loading...

we can then apply the multiplicitive property of limits again on the right limit to get:

loading...

since our limit variable h is not involved in the function in anyway we can simplify it further to:

loading...

now if we recall the definition of the derivitive:

loading...

loading...

we can substitute those in to our equaiton:

loading...

we have now completed our proof

Example 1: loading...

loading... Now you maybe wondering how this rule is applicable if we only have one function, but what we are able to do here, is define two functions that when multiplied together equate to loading.... In this case we can let loading... since loading.... We can now solve for the derivative loading.... Since the derivative of x is just one as it is a linear line with a slope(m value) of 1. We can substitute loading... and loading... with 1. To get loading... we can sub in the functions and simplify to get: loading..., which is exactly what the power rule would have given you. This example was purely for demonstration, I would recommend you read up on the power rule to take derivatives in the form of loading... as it is much faster than this.

Example 2: loading...

loading... Letâ€™s start by defining our functions: let loading... and loading... since loading... we can write loading... the derivative of loading... is loading..., the derivative of loading... is loading... (this is done with power rule) now we can substitute into loading... to get loading... now we can simplify: loading....

Example 3: loading...

loading... without the product rule we would be stuck using either the chain rule or expanding out, but with the product rule we can define loading... then loading... and since loading... we can differentiate loading... now we can substitute into loading... to get (loading... we can now simplify to get loading... we can validate our result using chain rule or by expanding before taking the derivative.

Example 4: loading...

loading... Now, you maybe thinking that we should let loading... and loading... but that is not allowed. Because by the definition of the product rule both functions must be differentials and loading... is not a differentiable function. so we cannot directly apply the product rule here. what we would have to do is create a piecewise function where it becomes loading... when loading... or loading... when loading... which as a piecewise would look like this: loading... then we can differentiate the two segments independently and use the definition of the derivitive to find the derivitve at their point of intersection.

Example 5: loading...

loading... This one is a little bit strange becuase using the product rule you will end up with loading... but if you leave it like this you will end up with a derivitive that is not defined at x=0, but in reality you need to simplify to loading... for your solution to be valid.

Pro tips:

- All functions must be differentiable for you to be able to use the product rule
- Can be used with any number of functions from 1 to any infinity
- can be used in place of the chain rule(with limited success)
- can be used in place of the power rule with limited success

THANKS FOR READING