@@JonnyMath Here's how you can prove it for real numbers in general. This assumes we already know the derivative of e^x, ln(x), and the chain rule. Given x^p, where p is a constant, we'd like to find d/dx x^p. Rewrite this in base e: e^(ln(x)*p) Differentiate with the chain rule: e^(ln(x)*p) * p/x Rewrite as a base x power: p*x^(p- 1) This shows that we get the power rule for positive x-values, exactly as we expect. You can extend this to show that it also works for negative x-values as well, by using complex numbers. When x is a negative number, the value of ln(x) = ln|x| + i*pi*(2*k + 1) Thus: e^(ln(x)*p) * p/x = e^((ln|x| + i*pi*(2*k + 1))*p) * p/x Expand with Euler's formula: e^((ln|x| + i*pi)*p) = |x|^p * [cos((2*k + 1)*p*pi) + i*sin((2*k + 1)*p*pi)] In order for this to have a real value, it must be the case that p is either an integer, or a fraction with an odd denominator. This way, (2*k + 1)*p will be an integer, and the sine term disappears, for a special case of the value of k. This is the version of the multivalued complex logarithm that matters. When this is the case, the derivative is: p*|x|^(p - 1) * cos((2*k + 1)*p*pi) The principal value, when (2*k + 1)*p = 1, will give us a cosine term of -1, which undoes the absolute value signs, and returns the original power rule: p*x^(p - 1) Where p is either a fraction with an odd denominator, or an integer.
@@JonnyMath you dont need to use the definition of derivative to prove it for real numbers, just change x^n to e^( ln(x) * n) and then use the rules for exponentials and the chain rule
But fortunately, now you know!!!😅🤣 I came up with it the other day (actually I guess more than two weeks ago!😅🤣) because I wanted a faster and simpler way to prove it because I don't remember the binomial theorem!!!😅🤣🤣🤣
Amazingly it has worked!!!😅🤣 And I guess you could even prove it for integers with negative exponents putting a minus sign in front of the exponent... I gotta try to do it!!!🤩🤩🤩
The induction rule works only for integers. The proof for all real numbers is a bit more involved I think.
Yes you're right!!! I don't really know how to prove it for real numbers... Gotta find out!!!😉🤗
First principles?
@@JonnyMath Here's how you can prove it for real numbers in general. This assumes we already know the derivative of e^x, ln(x), and the chain rule.
Given x^p, where p is a constant, we'd like to find d/dx x^p.
Rewrite this in base e:
e^(ln(x)*p)
Differentiate with the chain rule:
e^(ln(x)*p) * p/x
Rewrite as a base x power:
p*x^(p- 1)
This shows that we get the power rule for positive x-values, exactly as we expect.
You can extend this to show that it also works for negative x-values as well, by using complex numbers. When x is a negative number, the value of ln(x) = ln|x| + i*pi*(2*k + 1)
Thus:
e^(ln(x)*p) * p/x = e^((ln|x| + i*pi*(2*k + 1))*p) * p/x
Expand with Euler's formula:
e^((ln|x| + i*pi)*p) = |x|^p * [cos((2*k + 1)*p*pi) + i*sin((2*k + 1)*p*pi)]
In order for this to have a real value, it must be the case that p is either an integer, or a fraction with an odd denominator. This way, (2*k + 1)*p will be an integer, and the sine term disappears, for a special case of the value of k. This is the version of the multivalued complex logarithm that matters. When this is the case, the derivative is:
p*|x|^(p - 1) * cos((2*k + 1)*p*pi)
The principal value, when (2*k + 1)*p = 1, will give us a cosine term of -1, which undoes the absolute value signs, and returns the original power rule:
p*x^(p - 1)
Where p is either a fraction with an odd denominator, or an integer.
@@JonnyMath you dont need to use the definition of derivative to prove it for real numbers, just change x^n to e^( ln(x) * n) and then use the rules for exponentials and the chain rule
Nice!!!😉🤗
Have you ever thought about this PROOF???🤩🤩🤩 Let me know!!!⬇⬇⬇
No I have Not.
But fortunately, now you know!!!😅🤣 I came up with it the other day (actually I guess more than two weeks ago!😅🤣) because I wanted a faster and simpler way to prove it because I don't remember the binomial theorem!!!😅🤣🤣🤣
Amazingly it has worked!!!😅🤣 And I guess you could even prove it for integers with negative exponents putting a minus sign in front of the exponent... I gotta try to do it!!!🤩🤩🤩
Ahhh a fellow Minecraft playing - math enjoyer
It's been a while since I last played Minecraft!!!😅🤣