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,
Would you simply multiply the coefficient "3" into the expression? Like this:
f'(x)=3 * 4x^3
Is this right?(9 votes)
- 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).(10 votes)
- 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(7 votes)
- 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.(3 votes)
- How do you know when something is not a power function?(4 votes)
- 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.(5 votes)
- Why wasn't -x^-2 rewritten as -1/x^2 ? You're usually not supposed to leave negative exponents in answers.(3 votes)
- 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).(4 votes)
- 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.
The pattern I got was this:
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.(3 votes)
- 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)(3 votes)
- 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:
Thanks for your time, you who have read down to here.
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,
This was not accepted.
The "correct" answer was
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 = Infinityis 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.(3 votes)
- 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)(2 votes)
- At4:00, 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?(2 votes)
- can anyone provide me with the link of proof of power rule?(2 votes)
- IS d/dx same as dy/dx?(1 vote)
- 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.(2 votes)
- Can the power rule be used to find the derivative when there is a sum under the radical? Would I need to use the difference quotient and find the limit as h approaches 0 instead?
For example: f(x)=(7x+1)^1/2(1 vote)
- Use the power rule together with the chain rule.
𝑓(𝑥) = (7𝑥 + 1)^(1∕2) ⇒
⇒ 𝑓 '(𝑥) = (1∕2)(7𝑥 + 1)^(−1∕2)・𝑑∕𝑑𝑥(7𝑥 + 1) =
= (1∕2)(7𝑥 + 1)^(−1∕2)・7(2 votes)
- [Instructor] What we're going to do in this video is get some practice taking derivatives with the power rule. So let's say we take the derivative with respect to x of one over x. What is that going to be equal to? Pause this video and try to figure it out. So at first, you might say, "How does the power rule apply here?" The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to the n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is gonna be n times x to the, and then we decrement the exponent. So, n minus one. But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one. So, this is going to be the derivative with respect to x of x to the negative one. And now, this looks a lot more like what you might be used to, where this is going to be equal to, you take our exponent, bring it out front, so it's negative one, times x to the negative one minus one, negative one minus one. And so, this is going to be equal to negative x to the negative two, and we're done. Let's do another example. Let's say that we're told that f of x is equal to the cube root of x and we wanna figure out what f prime of x is equal to. Pause the video and see if you can figure it out again. Well, once again, you might say, "Hey, how do I take the derivative of something like this, "especially if my goal or if I'm thinking that maybe "the power rule might be useful?" And the idea is to rewrite this as an exponent, if you can rewrite the cube root as x to the 1/3 power. And so, the derivative, you take the 1/3, bring it out front, so it's 1/3 x to the 1/3 minus one power. And so, this is going to be 1/3 times x to the 1/3 minus one is negative 2/3, negative 2/3 power, and we are done. And hopefully through these examples, you're seeing that the power rule is incredibly powerful. You can tackle a far broader range of derivatives than you might have initially thought. Let's do another example, and I'll make this one really nice and hairy. Let's say we wanna figure out the derivative with respect to x of the cube root of x squared. What is this going to be? And actually, let's just not figure out what the derivative is, let's figure out the derivative at x equals eight. Pause this video again and see if you can figure that out. Well, what we're gonna do is first just figure out what this is and then we're going evaluate it at x equals eight. And the key thing to appreciate is this is the same thing, and we're just gonna do what we did up here as the derivative with respect to x. Instead of saying the cube root of x squared, we can say this is x squared to the 1/3 power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I can just take the product of the exponents. And so, this is gonna be x to the two times 1/3 power or to the 2/3 power. And now, this is just going to be equal to, I'll do it right over here, bring the 2/3 out front, 2/3 times x to the, what's 2/3 minus one? Well, that's 2/3 minus 3/3 or it would be negative 1/3 power. Now, we wanna know what happens at x equals eight, so let's just evaluate that. That's going to be 2/3 times x is equal to eight to the negative 1/3 power. Well, what's eight to the 1/3 power? Eight to the 1/3 power is going to be equal to two, and so, eight to the negative 1/3 power is 1/2. Actually, let me just do that step-by-step. So, this is going to be equal to 2/3 times, we could do it this way, one over eight to the 1/3 power. And so, this is just one over two, 2/3 times 1/2, well, that's just going to be equal to 1/3, and we're done.