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### Course: Differential Calculus (2017 edition)>Unit 6

Lesson 5: Radical functions differentiation (intro)

# Power rule review

Review your knowledge of the Power rule for derivatives and solve problems with it.

## What is the Power rule?

The Power rule tells us how to differentiate expressions of the form ${x}^{n}$ (in other words, expressions with $x$ raised to any power):
$\frac{d}{dx}{x}^{n}=n\cdot {x}^{n-1}$
Basically, you take the power and multiply it by the expression, then you reduce the power by $1$.
Want to learn more about the Power rule? Check out this video.

## Differentiating polynomials

The Power rule, along with the more basic differentiation rules, allows us to differentiate any polynomial. Consider, for example, the monomial $3{x}^{7}$. We can differentiate it as follows:
$\begin{array}{rl}\frac{d}{dx}\left[3{x}^{7}\right]& =3\frac{d}{dx}\left({x}^{7}\right)\phantom{\rule{1em}{0ex}}\text{Constant multiple rule}\\ \\ & =3\left(7{x}^{6}\right)\phantom{\rule{1em}{0ex}}\text{Power rule}\\ \\ & =21{x}^{6}\end{array}$
Problem 1
$f\left(x\right)={x}^{5}+2{x}^{3}-{x}^{2}$
${f}^{\prime }\left(x\right)=$

Want to try more problems like this? Check out this exercise.

## Differentiating negative powers

The Power rule also allows us to differentiate expressions like $\frac{1}{{x}^{2}}$, which is basically $x$ raised to a negative power. Consider this differentiation of $\frac{1}{{x}^{2}}$:
$\begin{array}{rl}\frac{d}{dx}\left(\frac{1}{{x}^{2}}\right)& =\frac{d}{dx}\left({x}^{-2}\right)\phantom{\rule{1em}{0ex}}\text{Rewrite as power}\\ \\ & =-2\cdot {x}^{-3}\phantom{\rule{1em}{0ex}}\text{Power rule}\\ \\ & =-\frac{2}{{x}^{3}}\phantom{\rule{1em}{0ex}}\text{Rewrite as fraction}\end{array}$
Problem 1
$\frac{d}{dx}\left(\frac{-2}{{x}^{4}}+\frac{1}{{x}^{3}}-x\right)=$

Want to try more problems like this? Check out this exercise.

## Differentiating fractional powers and radicals

The Power rule also allows us to differentiate expressions like $\sqrt{x}$ or ${x}^{{}^{\frac{2}{3}}}$. Consider this differentiation of $\sqrt{x}$:
$\begin{array}{rl}\frac{d}{dx}\sqrt{x}& =\frac{d}{dx}\left({x}^{{}^{\frac{1}{2}}}\right)\phantom{\rule{1em}{0ex}}\text{Rewrite as power}\\ \\ & =\frac{1}{2}\cdot {x}^{{}^{-\frac{1}{2}}}\phantom{\rule{1em}{0ex}}\text{Power rule}\\ \\ & =\frac{1}{2\sqrt{x}}\phantom{\rule{1em}{0ex}}\text{Rewrite as radical}\end{array}$
Problem 1
$f\left(x\right)=6{x}^{{}^{\frac{2}{3}}}$
${f}^{\prime }\left(x\right)=$

Want to try more problems like this? Check out these exercises:

## Want to join the conversation?

• What is a situation where you would need to use the constant multiple rule? Aren't all of the problems where you could use the constant multiple rule covered by the power rule?

For instance: If you wanted to differentiate 3x^2, you could simply multiply 2*3 and then subtract 1 from the exponent. However, Sal uses the constant multiple rule in many of his earlier examples to first pull out the 3 and then find the derivative of x^2. It seems like the constant multiple rule is an unnecessary step when finding derivatives.
(4 votes)
• The power rule essentially tells you how to differentiate x^n. It doesn't say anything about multiples of x^n, like 3x^2. So when you differentiate 3x^2, you really apply both the power rule and the constant multiple rule.

First, the constant multiple rule tells us that the derivative of 3x^2 is 3 times the derivative of x^2. Then, the power rule tells us that the derivative of x^2 is 2x. Put those two together, and you get that the derivative of 3x^2 is 6x.

Once you're experienced in differentiating various functions, you can "forget" the particular rules you use, and just do it fluently. So you quickly differentiate 3x^2 by multiplying 2 by 3 and reducing the 2nd power to 1. This doesn't mean you're not using the underlying rules, it just means you don't need to acknowledge them anymore.
(36 votes)
• I know that pi is a constant, then f(x) = pi*x^2 has the differentiation f'(x) = 2*pi*x. But what about 'e' (Euler)? its something pi? Its a constant? f(x) = e*x^2 with f'(x) = 2*e*x.
And, for what I use 'e'? And his differentiation: prime 'e'?
Thanks a lot!
(2 votes)
• e is a real number, just like π, √2, or 7. It's just a constant. All the derivatives you gave are correct.
(4 votes)
• (x^2 + 1)^2 how can i solve this using the power rule im stuck :((
(2 votes)
• To solve (x^2+1)^2, You have to multiply the power rule equation by its derivate.
For example, the ^2 on the outside will then move to the front of the function as part of the power rule.

So, 2(x^2+1) * D/DX (x^2+1). After that, you can find the derivate for each separate part of the function. So, d/dx of (x^2)=2x and d/dx of (1)=0. Finally, your answer would be 2(x^2+1)*(2x), where 2 is distributed to the 2x.

Thus your answer is f'(x)= 4x(x^2+1) which can be further reduced to 4x^3+4x.

Hope this helps ^u^
(1 vote)
• what about when it is in fraction form
(1 vote)
• Does it apply to constants? Because both constant and the term x have the same power but different derivatives
(1 vote)
• How was the power rule derived?
(2 votes)
• Can the exponent n be zero? The function x^0 is 1 for all x, except maybe when x=0. Its derivative is 0 because is a constant. And the power rule says its derivative is 0•x^(0-1) which is 0 for all x values, but it's not defined when x=0.
(1 vote)
• Can you use the power rule for terms like (5x + 1)? If this were, say, raised to the 4th power would you be able to use the power rule, or would you have to use the chain rule?
(1 vote)
• Why can't I use power rule, when a constant is raised to power?
For Example: a^x
(1 vote)
• Can you please help me solve F(t)=(5t-8)^2?
(0 votes)