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

Lesson 5: Applying the power rule

# Power rule (with rewriting the expression)

Discover how the power rule helps us find derivatives of functions like 1/x, ∛x, or ∛x². By rewriting these functions as xⁿ, where n is a negative or fractional exponent, we can apply the power rule to calculate their derivatives with ease.

## 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?
• Yes, you're right. You take the exponent (4) down, multiply the 4 into the original expression, and decrement the exponent by 1 (after differentiation the exponent is 3).
• 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
• I think it might depend on which playlist you are watching this on (I think playlist is the right word for it.) Have you seen f'(x)? because it's basically another way of writing that.
• How do you know when something is not a power function?
• The only functions that are power functions are a variable brought to some number power. if it is anything else it is not a pwer function. So exponential, logarithmic, trig functions, those are all NOT power functions.
• IS d/dx same as dy/dx?
• Not really. d/dx is an operator while dy/dx is an operation. d/dx simply says "the derivative with respect to x" but dy/dx says "the derivative with respect to x of y". So, d/dx in itself has no meaning, as it gives no information on what you're differentiating.
• Why wasn't -x^-2 rewritten as -1/x^2 ? You're usually not supposed to leave negative exponents in answers.
• I don't know exactly, why are you not supposed to leave negative exponents in answers?
I think that your right, the answer you gave was more simplified, but there both right answers (I think).
• 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.
• When you said: "d/dx 1/x^a = d/dx x^-a = -ax^(-a-1) = 1/-ax^(a+1)", you made the error of saying "-ax^(-a-1) = 1/-ax^(a+1)". This is incorrect. The correct fraction is: "-ax^(-a-1) = -a/x^(a+1)". So the correct formula is d/dx 1/x^a = -a/x^(a+1)