# Product Rule Calculator + Online Solver With Free Steps

The Product Rule Calculator is used to solve Product Rule Problems as they cannot be solved using traditional techniques for calculating the derivative. Product Rule is a formula derived from the definition of the derivative itself, and it is very useful in the world of Calculus.

As most problems Engineers and Mathematicians face daily mostly include multiple different functions having different operations applied amongst them. And this Product Rule is one of a series of Rules which are derived to cater to such special case scenarios.

## What is a Product Rule Calculator?

A Product Rule Calculator is an online calculator which is designed to solve differentiation problems in which the expression is a product of two differentiable functions.

These differentiable functions, therefore, need to be solved using the Product Rule, a formula that has been derived especially for problems of such sort.

Thus, this is a unique calculator with its roots in Calculus and Engineering. And it can solve these complex problems inside your browser with no requirements of its own. You can simply place your differential expressions in it and get solutions.

## How to Use the Product Rule Calculator?

To use the Product Rule Calculator, you must first have a problem which you may want to find the differential that also fits the criteria for the Product Rule Calculator. This means that it must have a couple of functions multiplied together for the Product Rule to be used.

Once acquired, this expression can then be transformed into the correct format for the Calculator to be able to read it properly. After doing that you can simply place this Differential Equation into the input box, and watch the magic happen.

Now, to get the best results from your calculator experience, follow the step-by-step guide given below:

### Step 1

First, you must have a function with differential applied to it, and in the correct format for the calculator to read.

### Step 2

Then you can simply enter this differential equation into the input box labeled: “Enter the function =”.

### Step 3

Following entering the product of functions, you are to press the button labeled “Submit” as it will provide you with your desired results in a new window.

### Step 4

Finally, you can choose to either close this new window or keep using it if you intend to solve more problems of similar nature.

It may be important to note that this calculator can only solve problems with two functions forming a product. As the calculations become far more complex going into a higher number of constituting functions.

## How Does the Product Rule Calculator Work?

The Product Rules Calculator works by solving the derivative for the product of two functions using the Product Rule for differentiation. It is necessary just to run the input functions through a bunch of first-order Derivative Calculations and place the results in a formula.

Now, before we try to understand where this formula comes from, we must go into detail about the Product Rule itself.

### Product Rule

The rule is also called Leibniz Rule after the renowned mathematician, who derived it. This rule is of great significance in the world of Calculus. The Product Rule is a formula to solve the calculus involved in the Differentiation of an expression involving a product of two differentiable functions.

It can be expressed in its simplified form as follows:

For a function of x, f(x) the definition is constituted by two functions u(x), and v(x).

f(x) = u(x) . v(x)

And differentiating this function according to the Product Rule looks like this:

f'(x) = [v(x) . u'(x) + u(x) . v'(x)]

It is one of the many rules derived for different types of operations occurring between differentiable functions constituting one in the process themselves.

### Product Rule Derivation

Now to derive this equation called Product Rule, we must first go back to the basic definition of a derivative of a function h(x). The derivative of this function is given below:

$\frac{dy}{dx} = \frac{h(x + dx) – h(x)}{dx}$

Now, we assume that there is a function h(x) which is described as: h(x) = f(x) . g(x). Thus, this function h(x) constitutes of two functions Multiplied Together i.e., f(x), and g(x).

Let’s combine these both now:

$h'(x) = \lim_{dx\to0} \frac{h(x + dx) – h(x)}{dx}$

$= \lim_{dx\to0} \frac{f(x + dx)g(x + dx) – f(x)g(x)}{dx}$

$= \lim_{dx\to0} \frac{[f(x + dx) – f(x)]g(x + dx)}{dx} + \lim_{dx\to0} \frac{[g(x + dx) – g(x)]f(x)}{dx}$

$= \bigg ( \lim_{dx \to 0} \frac{f(x + dx) – f(x)}{dx} \bigg ) \bigg ( \lim_{dx \to 0} g(x + dx) \bigg) + \bigg ( \lim_{dx \to 0} f(x) \bigg ) \bigg ( \lim_{dx \to 0} \frac{g(x + dx) – g(x)}{dx} \bigg)$

$= g(x) \bigg ( \lim_{dx \to 0} \frac{f(x + dx) – f(x)}{dx} \bigg ) + f(x) \bigg ( \lim_{dx \to 0} \frac{g(x + dx) – g(x)}{dx} \bigg)$

$= \begin{matrix} Where, & f'(x) = \lim_{dx \to 0} \frac{f(x + dx) – f(x)}{dx} & and & g'(x) = \lim_{dx \to 0} \frac{g(x + dx) – g(x)}{dx} \end{matrix}$

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

Therefore, we have extracted the Product Rule formula by deriving it from the differential definition.

### Deriving Product Rule From Chain Rule

We have already derived the Product Rule from the differentiation of a function’s definition, but we can also use the Chain Rule to describe the validity of the Product Rule. Here, we will take up the Product Rule as an unusual case of the Chain Rule, where the function h(x) is expressed as:

h(x) = f(x) . g(x)

Now, applying the derivative on this expression can look like this:

$\frac{d}{dx} h(x) = \frac{d}{dx} f \cdot g = [\frac{d}{df} (fg)] [\frac{df}{dx}] + [\frac{d}{dg} (fg)] [\frac{dg}{dx}] = g(\frac{df}{dx}) + f(\frac{dg}{dx})$

Finally, we have the Product Rule formula again, this time derived using the Chain Rule Principle of differentiation.

### Differentiation of a Product With More Functions Than Two

It may be important to look at a Differentiation of more than two functions being multiplied together, as things may change slightly moving to a larger number of functions. This can be tackled by the same Product Rule Formula so there is nothing to worry about. So, let’s see what happens for a function of that nature:

$\frac{d(uvw)}{dx} = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx} \cdot \frac{d(uvw)}{dx} = \frac{du}{dx} vw + u \frac{dv}{dx} w + uv \frac{dw}{dx}$

This is an example of 3 functions multiplied together, and this goes to show us a pattern for a possible solution for the n number of functions here.

## Solved Examples

Now that we have learned a lot about how the Product Rule was derived, and how it is used on a theoretical level. Let’s go further and watch how it is used to solve a problem where it is needed. Here are a few examples to observe where we are solving two function problems using the Product Rule.

### Example 1

Consider the given function:

f(x) = x . log x

Solve the first-order derivative for this function using Product Rule.

### Solution

We begin by first separating the different parts of this function into their respective representations. This is done here:

f(x) = u(x) . v(x)

$\begin{matrix}u(x) = x, & v(x) = \log x \end{matrix}$

Now we apply first derivatives on these u and v snippets of the original function. This is carried out as follows:

$\begin{matrix}u'(x) = \frac{d}{dx} (x) = 1, & v'(x) = \frac{d}{dx} (\log x) = \frac{1}{x} \end{matrix}$

Once through with the calculation of the first order derivatives, we move forward to introducing the Product Rule Formula as given below:

f'(x) = [v(x) . u'(x) + u(x) . v'(x)]

Placing in the values calculated above will give us the end result i.e., solution to the derivative of the given product of two functions.

f'(x) = log x . 1 + x . $\frac{1}{x}$ = log x + 1

### Example 2

Consider the combination of functions given as:

$f(x) = (1 – x^3) e^{2x}$

Solve for the first-order differential of this expression using the Product Rule of Differentiation.

### Solution

We start off by rearranging the given equation in terms of the functions it is made from. This can be done as follows:

f(x) = u(x) . v(x)

$\begin{matrix}u(x) = (1 – x^3), & v(x) = e^{2x} \end{matrix}$

Here, we have u and v, both representing the constituents of the original f(x). Now, we must apply derivative on these constituting functions and get u’, and v’. This done here:

$\begin{matrix}u'(x) = \frac{d}{dx} (1 – x^3) = -3x^2, & v'(x) = \frac{d}{dx} (e^{2x}) = 2e^{2x} \end{matrix}$

Now, we have all the required pieces to build up to the result. We bring in the formula for the Product Rule for the derivative of multiplying values.

f'(x) = [v(x) . u'(x) + u(x) . v'(x)]

Finally, we conclude by putting in the values we have calculated above and therefore finding the solution to our problem as follows:

$f'(x) = e^{2x}\cdot -3x^2 + (1 – x^3) \cdot 2e^{2x} = e^{2x}(2 – 3x^2 – 2x^3)$