# 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 $v(x)$.

\[f(x) = u(x) \cdot v(x)\]

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

\[f'(x) = [v(x) \cdot u'(x) + u(x) \cdot v'(x)]\]

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: $h(x) = f(x) \cdot g(x)$. Thus, this function $h(x)$ constitutes of two functions **Multiplied Together** i.e., $f(x)$, and $g(x)$.

Let’s combine these both now:

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

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

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

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

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

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

\[ h'(x) = g(x) \cdot f'(x) + g'(x) \cdot f(x)\]

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:

\[h(x) = f(x) \cdot g(x)\]

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:

\[f(x) = x \cdot \log x\]

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:

\[f(x) = u(x) \cdot v(x)\]

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

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:

\[f'(x) = [v(x) \cdot u'(x) + u(x) \cdot v'(x)]\]

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.

\[f'(x) = log x \cdot 1 + x \cdot \frac{1}{x} = \log x + 1\]

### 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:

\[f(x) = u(x) \cdot v(x)\]

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

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.

\[f'(x) = [v(x) \cdot u'(x) + u(x) \cdot v'(x)]\]

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

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