AP®︎/College Calculus AB
- Proof: Differentiability implies continuity
- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof
Proof of power rule for square root function
Sal gives a proof of the power rule for the specific case where n=½ (i.e. for the derivative of √x). Created by Sal Khan.
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- I thought we could only directly substitute the value x is approaching (i.e. lim delta x→0) for polynomial and other continous functions like sin x and cos x. Since the function, square root of x, isn't continous everywhere (it is only continous from 0 to positive infinty), is it right to just plug in the value: delta x = 0 at the end of this proof? Please correct me if I'm wrong.(33 votes)
- It is okay to directly substitute the value x is approaching anytime you can simplify so that you would not be dividing by zero.
Notice though that where the function was continuous restricts where the derivative exists! This is because the derivative of a function is not defined wherever the function is not continuous.
The square root of x is continuous on the semi-closed interval [0, infinity), while its derivative exists only on the open interval (0, infinity).(37 votes)
- Can't this just be proved by plugging into dx/dy(x^n), which we just proved in the last video?(5 votes)
- If you want to prove the general case d/dx(x^r)=rx^(r-1) for any real number r, then you can use the chain rule on x^r = e^(ln(x)r). By the fact that d/dxe^x=e^x and d/dxln(x)=1/x one gets d/dx x^r = d/dx e^(ln(x)r) = e^(ln(x)r) d/dx ln(x)r = x^r 1/x r = r x^(r-1). QED(15 votes)
- I am having a lot of trouble with a problem:
My book tells me that I then flip this an it becomes:
But I am baffled as to why this happens. Somebody please help me!(4 votes)
- You may also multiply the numerator and denominator by sqrt(x), which gives sqrt(x)/x as (sqrt(x)^2)/(x sqrt(x)). Since the original expression is restricted to x>0 (so that there's no need to consider the absolute value), this is (x))/(x sqrt(x)). The x factors cancel, giving 1/sqrt(x)(6 votes)
- How would you evaluate:
f'(x) = abs(x)
And is Leibniz's notation considered less accurate than d/dx(4 votes)
- Consider the abs(x) = (x^2)^(1/2) or the square root of the square of x. Here, you want to perform the "Chain Rule" or u-substitution. Let u = x^2, u' = 2x.
Therefore, d/dx |x| = d/dx ((x^2)^(1/2)) = (1/2) ((x^2)^(-1/2)) (2x) = x/|x|
To generalize, let f(x) = | g(x) |
Then, going through the process by setting u = g(x),
d/dx |g(x)| = d/dx ((u^2)^(1/2)) = (1/2) ((u^2)^(-1/2)) (2uu') = (u')u/|u| = g'(x) * g(x) / |g(x)|(3 votes)
I try to differentiate this but it does not work. Used the chain rule.(3 votes)
- Try separating each individual part of the function.
First: 2x. Remember that if you have a line with the basic y=mx+b format, the derivative of that equation equals the coefficient of x, because that number describes the slope.
Second: 100. 100 is a constant, which you hopefully know equal zero when derived.
Third: sqrt((x^2)-243). If you just derive sqrt(x), you get 1/(2*sqrt(x)). Now for the chain rule, derive the square root part of the little segment we have here, treating the sqrt((x^2)-243) as the sqrt(x) in the derivative of sqrt(x), stated above. Multiply what you got there by the derivative of the inside, which is (x^2)-243. The constant cancels out to zero and the exponential turns into a nice even expression (which I will not state because you learn better if you figure it out yourself).
Mush that all together and simplify a tad, and you should get the answer. Hope this helps!(4 votes)
- why do we take sqrt(x+deltax) instead of sqrt(x) ?(3 votes)
- The definition of a derivative is the limit as deltax approaches zero of [f(x+deltax) - f(x)]/ deltax. So, since our function is sqrt(x), we plug in (x+deltax) for x and get sqrt(x+deltax). We do this because we want to find the slope as the interval in which we are taking the slope in approaches zero. By doing this "limit quotient" (as it is often called), we find deltay/deltax as deltax approaches zero, and therefore find the instantaneous rate of change.(4 votes)
- In finding limits, why do we not just directly put in the value 0 instead of doing complicated simplifications? Is there any good reason?(3 votes)
- If you put 0 in for "delta x" in the definition of a derivative, you will always get does not exist. If you mean that we should input the value of the limit we are trying to find, there may be a hole in the function that would result in another value.(3 votes)
- What if there's a trig function inside the square root? Do I chain rule? Say y=sqrt(4+7tan(x))(3 votes)
- Yes. If you apply the chain rule. The derivative of y=sqrt(4+7tan(x)) is 1/(2*sqrt(4+7tan(x)))(the derivative of the outside function)*7sec(x)^2(the derivative of the inside function).(3 votes)
- how do you find the derivative for y = (5x-5)^1/2(3 votes)
- You would have to apply the chain rule.
d/dx (5x-5)^1/2 = (1/2)(5x-5)^-1/2 * 5 = 5/(2 * sqrt(5x-5))(3 votes)
- please help me with this.. I want to find the solution and answer in finding the derivative of this.
y= square root of 1 + sqrt of 1 + sqrt of x. thank you(3 votes)
So I've been requested to do the proof of the derivative of the square root of x, so I thought I would do a quick video on the proof of the derivative of the square root of x. So we know from the definition of a derivative that the derivative of the function square root of x, that is equal to-- let me switch colors, just for a variety-- that's equal to the limit as delta x approaches 0. And you know, some people say h approaches 0, or d approaches 0. I just use delta x. So the change in x over 0. And then we say f of x plus delta x, so in this case this is f of x. So it's the square root of x plus delta x minus f of x, which in this case it's square root of x. All of that over the change in x, over delta x. Right now when I look at that, there's not much simplification I can do to make this come out with something meaningful. I'm going to multiply the numerator and the denominator by the conjugate of the numerator is what I mean by that. Let me rewrite it. Limit is delta x approaching 0-- I'm just rewriting what I have here. So I said the square root of x plus delta x minus square root of x. All of that over delta x. And I'm going to multiply that-- after switching colors-- times square root of x plus delta x plus the square root of x, over the square root of x plus delta x plus the square root of x. This is just 1, so I could of course multiply that times-- if we assume that x and delta x aren't both 0, this is a defined number and this will be 1. And we can do that. This is 1/1, we're just multiplying it times this equation, and we get limit as delta x approaches 0. This is a minus b times a plus b. Let me do little aside here. Let me say a plus b times a minus b is equal to a squared minus b squared. So this is a plus b times a minus b. So it's going to be equal to a squared. So what's this quantity squared or this quantity squared, either one, these are my a's. Well it's just going to be x plus delta x. So we get x plus delta x. And then what's b squared? So minus square root of x is b in this analogy. So square root of x squared is just x. And all of that over delta x times square root of x plus delta x plus the square root of x. Let's see what simplification we can do. Well we have an x and then a minus x, so those cancel out. x minus x. And then we're left in the numerator and the denominator, all we have is a delta x here and a delta x here, so let's divide the numerator and the denominator by delta x. So this goes to 1, this goes to 1. And so this equals the limit-- I'll write smaller, because I'm running out of space-- limit as delta x approaches 0 of 1 over. And of course we can only do this assuming that delta-- well, we're dividing by delta x to begin with, so we know it's not 0, it's just approaching zero. So we get square root of x plus delta x plus the square root of x. And now we can just directly take the limit as it approaches 0. We can just set delta x as equal to 0. That's what it's approaching. So then that equals one over the square root of x. Right, delta x is 0, so we can ignore that. We could take the limit all the way to 0. And then this is of course just a square root of x here plus the square root of x, and that equals 1 over 2 square root of x. And that equals 1/2x to the negative 1/2. So we just proved that x to the 1/2 power, the derivative of it is 1/2x to the negative 1/2, and so it is consistent with the general property that the derivative of-- oh I don't know-- the derivative of x to the n is equal to nx to the n minus 1, even in this case where the n was 1/2. Well hopefully that's satisfying. I didn't prove it for all fractions but this is a start. This is a common one you see, square root of x, and it's hopefully not too complicated for proof. I will see you in future videos.