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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.
• 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).
• Can't this just be proved by plugging into dx/dy(x^n), which we just proved in the last video?
• 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
• I am having a lot of trouble with a problem:
sqrt(x)/x

My book tells me that I then flip this an it becomes:
1/sqrt(x)

• 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)
• How would you evaluate:
f'(x) = abs(x)
And is Leibniz's notation considered less accurate than d/dx
• 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)|
• d/dx(2x+100-sqrt((x^2)-243))
I try to differentiate this but it does not work. Used the chain rule.
• 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!
• why do we take sqrt(x+deltax) instead of sqrt(x) ?
• 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.
• In finding limits, why do we not just directly put in the value 0 instead of doing complicated simplifications? Is there any good reason?
• 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.
• What if there's a trig function inside the square root? Do I chain rule? Say y=sqrt(4+7tan(x))
• 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).