**Math Calculators** ▶ Product Rule Derivative Calculator

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**Table of Content**

An online product rule derivative calculator helps you to determine the derivative of a function that is composed of smaller differentiable functions. This calculator uses the product rule of differentiation to simplify your problem precisely. This content is packed with a whole radical information about the product rule.

Read on!

In product rule calculus, we use the multiplication rule of derivatives when two or more functions are getting multiplied.

If we have two functions **f(x)** and **g(x)**, then the product rule states that:

**“ f(x) times the derivative of g(x) plus g(x) times the derivative of f(x)”**

Suppose that we have two functions** f(x)** and** g(x)** that are differentiable. The derivative product rule formula for these functions is as follows:

$$ \frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)} $$

Apart from using formula for manual calculations, use online product rule derivative calculator for free to find derivative of two product functions.

You can simplify the product of two functions using the basic derivative multiplication rule.

Let us solve a couple of examples.

**Example # 01:**

Differentiate the following function **w.r t x.**

$$ h\left( x \right) = \left( {6{x^2} – x} \right)\left( {1 – 30x} \right) $$

**Solution:**

The given function is:

$$ h\left( x \right) = \left( {6{x^2} – x} \right)\left( {1 – 30x} \right) $$

As we know that the multiplication derivative rule is as follows:

$$ \frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)} $$

Here we have separate functions according to the formula as follows:

$$ f(x) = \left( {6{x^2} – x} \right) $$

$$ g(x) = \left( {1 – 30x} \right) $$

Now, calculating the derivative of f(x) w.r.t x:

$$ \frac{d}{d x} f(x) $$

$$ =\frac{d}{d x} \left(6 x^{2} – x\right) $$

$$ \frac{d}{d x} \left(6 x^{2} – x\right) = \left(12x – 1\right) $$

$$ \frac{d}{d x} g(x) $$

$$ =\frac{d}{d x} \left(1 – 30 x\right) $$

$$ \frac{d}{d x} \left(1 – 30 x\right) = -30 $$

(For step-by-step calculation of derivative, click derivative calculator)

Now, according to the multiplication rule for derivatives:

$$ \frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)} $$

Putting derivatives in the formula to get the final answer.

$$ \left(6 x^{2} – x\right) (-30) + \left(1 – 30 x\right) \left(12 x – 1\right) $$

$$ – 180 x^{2} + 30 x + \left(1 – 30 x\right) \left(12 x – 1\right) $$

after simplifying , we get:

$$ – 180 x^{2} + 30 x + (1)(12x) – (1)(-1) -(30x)(12x) + (-30x)(-1) $$

$$ – 540 x^{2} + 72 x – 1 $$

So we have:

$$ h\left( x \right) = \left( {6{x^2} – x} \right)\left( {1 – 30x} \right) = – 540 x^{2} + 72 x – 1 $$

Which is the required answer.

Also, our free online product rule derivative calculator evaluates the given functions more accurately and instantly.

**Example # 02:**

Differentiate the following function according to the multiplication rule derivatives w.r t the variable z.

$$ (z^2)^{1/3} *(2z-z^2) $$

**Solution:**

As the product rule is given as:

We define the functions individually:

$$ f(x) = (z^{2})^{1/3} $$

$$ g(x) = \left(2z – z^{2}\right) $$

Taking derivatives of both the functions:

$$\frac{d}{d z} f(x) $$

$$ =\frac{d}{d z} \sqrt[3]{z^{2}} $$

$$ \frac{d}{d z} \sqrt[3]{z^{2}} = \frac{2 \sqrt[3]{z^{2}}}{3 z} $$

$$ \frac{d}{d z} g(x) $$

$$ =\frac{d}{d z} \left(- z^{2} + 2 z\right) $$

$$ \frac{d}{d z} \left(- z^{2} + 2 z\right) = 2 – 2z $$

(For step-by-step calculation of derivative, click derivative calculator)

Following the product rule for derivatives:

Putting values for derivatives of each function and simplifying:

$$ \frac{2 \left(5 – 4 z\right) \sqrt[3]{z^{2}}}{3} $$

Which is our required answer.

To evaluate the derivative of two or more functions that are multiplying, you need to follow a simple guide as follows:

**Input:**

- Enter the given function in the equation menu that is supported by various functions like log, sqrt, ln, sin, cos and tan etc.
- Select the variable w.r.t which you wish to determine the derivative of the function that is given. The available variables are a, b, c, d, x, y, z or n.
- Select limit of differentiation that can’t be exceeded than 5.
- Click ‘calculate’

**Output:**

Our free product rule derivatives calculator calculates:

- Overall derivative of the function by following the product rule.
- Simplifies your problem in a proper manner.
- Step by step calculations to better understand the problem structure.

The product rule of exponents states that:

**“When we multiply the exponential expressions having the same base, we add their exponents”**

$$ \left(a^{m}*a^{n}\right) $$

$$ =a^{m+n} $$

**For example:**

$$ \left(a^{5}*a^{8}\right) $$

$$ =a^{5+8} $$

$$= a^{13} $$

Yes, you can do so. All you need to do is consider derivatives for each new function in the expression and add them to get the final answer.

The natural logarithm **(ln)** is defined only for **x>0**. That is why natural log of zero is undefined.

**ln(0) = ∞**

As we know that:

**log(e) = 1**. So, we have:

**dy / dx = 0**

The reason is that we know the derivative of any constant term is always zero.

The product rule of differentiation has strong applications in the field of calculus and engineering science. Mathematicians make a vast use of free online product rule derivative calculator to differentiate the complex functions at a given point. This calculator helps professionals and students on the same scale to get a generic solution to their problems quickly.

From the source of wikipedia: Chain rule, Smooth infinitesimal analysis, Quotient rule, Derivatives of inverse functions.

From the source of khan academy: Quotient rule, Differentiate quotients, Quotient rule with table, Differentiating rational functions.

From the source of lumen learning: Derivatives and Rates of Change, The Derivative as a Function, Differentiation Rules, Derivatives of Trigonometric Functions, Implicit Differentiation, Higher Derivatives.