How To Find The Derivative Of 3 Products

Much of calculus and finding derivatives is almost determining which rule applies to which instance. The product rule, simply put, is applied when your office is the product of two other functions.

Formula for the product rule. For fg, write fg + fg, then take the derivative of the first and the derivative of the second to get f prime g plus f g prime.
In this guide, we will expect at how to remember the product rule, how to recognize when it should be used, and finally, how to use it.

Table of contents:

Remembering the product dominion
Examples
Hint: Watch for shortcuts

Remembering the product rule

There is an piece of cake trick to remembering this important rule: write the product out twice (adding the ii terms), and then discover the derivative of the first term in the showtime product and the derivative of the second term in the 2d production.
the-product-rule-steps

Examples

The easiest way to empathise when this applies and how to use it is to look at some examples. In each case, pay special attention to how nosotros identify that we are looking at a product of two functions.

Example

Find the derivative of the office.

\(f(ten) = ten^iv\ln(x)\)

Solution

This function is the product of 2 simpler functions: \(ten^4\) and \(\ln(ten)\). Therefore, nosotros tin can apply the product rule to find its derivative.

Write the product out twice, and put a prime on the first and a prime on the 2d:

\(\left(f(x)\right)^{\prime number} = \left(x^iv\right)^{\prime number}\ln(ten) + x^iv\left(\ln(x)\right)^{\prime}\)

Accept the derivatives using the rule for each function. Remember that for \(x^4\), you will utilize the power dominion and that the derivative of \(\ln(x)\) is \(\dfrac{1}{x}\).

\(\begin{align}\left(f(10)\right)^{\prime} &= \left(x^4\right)^{\prime}\ln(x) + x^four\left(\ln(x)\right)^{\prime number}\\ &= \left(4x^3\right)\ln(x) + ten^iv\left(\dfrac{1}{10}\right)\end{marshal}\)

Simplify, if possible.

\(\begin{align}&= \left(4x^3\right)\ln(10) + x^3\\ &= \boxed{ten^three\left(4\ln(x) + one\right)}\end{marshal}\)

Either of the last ii lines can be used equally a final answer, but the last one looks a little nicer and is probably going to be preferred by your teacher if you are currently taking calculus!

Permit's look at another case to make sure you got the basics downwards.

Example

Find the derivative of the function.

\(y = 2xe^x\)

This is the production of \(2x\) and \(e^x\), so we apply the production dominion. Remember that the derivative of \(2x\) is 2 and the derivative of \(east^x\) is \(eastward^x\).

\(\begin{align}y^{\prime number} &= \left(2x\correct)^{\prime number}east^x + 2x\left(e^x\right)^{\prime}\\ & = 2e^x + 2xe^10\\ &= \boxed{2e^x\left(i + x\right)}\stop{marshal}\)

As you can run into, with product rule problems, you are really just changing the derivative question into ii simpler questions.

Hint: Watch for shortcuts

This hint could likewise be called "now that you lot know the product dominion, don't go blindly applying it". To sympathise what that means, consider the post-obit function:

\(y = \left(x+4\right)\left(x+1\right)\)

This is a product of \(x+4\) and \(x+i\), and so if we want to detect the derivative, we should employ the product rule, correct?

It's true – you could use that. Even so, it is more piece of work than recognizing that, by FOILing you lot go:

\(y = \left(10+4\right)\left(10+1\right) = x^2+5x + 4\)

Then by applying the power dominion you have:

\(y^\prime = \left(x^2+5x + four\right)^{\prime number} = 2x + 5\)

This is the kind of thing you want to learn to discover. There are many issues where you lot tin salvage yourself some calculus workby simplifying ahead of time. In the examples before, however, that wasn't possible, and then the product rule was the best approach.

Summary

The product rule is used to find the derivative of any function that is the product of two other functions. The quickest manner to remember information technology is by thinking of the general pattern it follows: "write the production out twice, prime on 1st, prime on 2nd".

Continue studying derivatives

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