If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 2

Lesson 5: Applying the power rule

# Power rule (with rewriting the expression)

We can use the power rule to find the derivatives of functions like 1/x, ∛x, or ∛x². To do that, we first need to rewrite those functions as xⁿ, where n would be negative or a fraction.

## Want to join the conversation?

• How about when a coefficient is in the function? For example,
f(x)=3x^4
Would you simply multiply the coefficient "3" into the expression? Like this:
f'(x)=3 * 4x^3
f'(x)= 12x^3
Is this right? • Maybe I wasn't following closely, but it seems to me that this is the first time this notation is introduced, without explaining it: d/dx • How do you know when something is not a power function? • Why wasn't -x^-2 rewritten as -1/x^2 ? You're usually not supposed to leave negative exponents in answers. • Hey, so I was just toying around(at 3 A.M. like all of us do) and I found an interesting pattern, somewhat of a general formula for a tangent to a curve.
Let there be a function f(a)
Equation for a tangent to f(a):
f(x) = f'(a).x + (f(a)-a.f'(a))
where you can vary a and get the equation for a tangent to f(a) at a.
I tried this for many functions and it worked for all.
Is this true for all functions. If so can someone prove this rigorously?Thanks. • Let 𝑓(𝑥) be a function differentiable at 𝑥 = 𝑎.

The tangent line to 𝑓(𝑥) at 𝑥 = 𝑎 will be on the form
𝑦 = 𝑚𝑥 + 𝑏

By definition of derivative, 𝑚 = 𝑓 '(𝑎)

Also, we know that the tangent line passes through (𝑎, 𝑓(𝑎)), which gives us
𝑏 = 𝑓(𝑎) − 𝑚𝑎 = 𝑓(𝑎) − 𝑓 '(𝑎) ∙ 𝑎

So, we can write the tangent line to 𝑓(𝑥) at 𝑥 = 𝑎 as
𝑦 = 𝑓 '(𝑎) ∙ 𝑥 + 𝑓(𝑎) − 𝑓 '(𝑎) ∙ 𝑎 = 𝑓 '(𝑎) ∙ (𝑥 − 𝑎) + 𝑓(𝑎)
• Here's an insight that I had while doodling around with the power rule:

`d/dx 1/x^1 = d/dx x^-1 = -1x^-2 = 1/(-1x^2 )`
`d/dx 1/x^0 = d/dx x^-0 = 0x^-1 = 1/( 0x^1 )`
`d/dx 1/x^-2 = d/dx x^2 = 2x^1 = 1/( 2x^-1)`

I believe I've found an "Inverse power rule", where
`d/dx 1/x^a = 1/(-ax^(a+1))`

I feel like this is a true pattern, but can someone tell me:
` `1. Is this even a true pattern?
` `2. Is this just something completely obvious and natural or is it really as awesome as I think?
` `3. Does something like this already exist?
` `4. Does it work for all numbers?

Also, please tell me if some of my calculations are wrong and I'll correct them, I'm regularly online.

EDIT: This kinda makes sense, because:
`d/dx 1/x^a = d/dx x^-a = -ax^(-a-1) = 1/-ax^(a+1)`

EDIT 2: Do not try this on KA. They gave me a problem like this:
f(x) = 1/(x^10)
f'(x) = ?

I put in, according to my newly-acquired formula,
`1/-10x^(11)`
This was not accepted.
`-10x^(-11)`

My answer actually WAS CORRECT, but it wasn't in the form they accept.

So, for the KA problems, the formula you need is
`d/dx 1/x^a = -ax^(-a-1)`

EDIT 3: @teghsingh04 made a great point, saying that one of my calculations above, `d/dx 1/x^0 = d/dx x^-0 = 0x^-1 = 1/( 0x^1 )` must be wrong, because `1/(0x^1) = 1/0 = Infinity` is not equal to the common sense solution, `d/dx 1/x^0 = d/dx x^-0 = d/dx 1 = 0`. So, I think something doesn't work when there's a zero power in the denominator. • At , Sal expresses 8^-1/3 as 1/(8^3). In one of the practice questions, x^-8 solved at x=-1 is said to be -1. I can only see that happening if we exclude the negative when we raise -1 to an exponent (treating it as -(1^8) instead of as (-1)^8. Is this usual for the derivative power rule? A typo? • can anyone provide me with the link of proof of power rule?   