Derivative proof with a fractional exponent

[mathwonk]

1) the binomial theorem for fractional exponents is due to Newton.

2) if you want very rigorous proofs of every detail in calculus, you should not be reading calculus made easy, which is an intuitive presentation.

3) fractional derivatives are treated by riemann, in the complex case, by the cauchy integral formula, which changes the order of the derivative, into the order of an exponent under the integral sign.

i.e. since one can integrate fractional exponents, one can take fractional derivatives.

 

[DavidWhitbeck]

The binomial theorem says nothing about fractional exponents.

That is simply not true. Just use the gamma function in place of factorials, and the rest is the same.

 

[mathwonk]

The difference is semantic, since people who know only a little math, may think the phrase "binomial theorem" refers only to the one they learned for integer powers.

 

[Vid]

https://www.amazon.com/dp/9810234570/?tag=pfamazon01-20

For anyone interested especially with amazon's nifty search inside feature.

 

[Vid]

There's also a $10 dover book for those really interested.

https://www.amazon.com/dp/0486450015/?tag=pfamazon01-20

 

[robert Ihnot]

Chemical Penguin: $$= \sqrt {x} + \frac {1} {2} \frac {dx} {\sqrt {x}} - \frac {1} {8} \frac {dx^2} {x \sqrt {x}} +...$$ "terms with higher powers of dx"

There is some confusion with the above as written since higher powers of dx refer, for example, to (dx)^n, which is to say we are only taking exponents, not derivatives.

What is involved is sometimes (Wikipedia) referred to as Newton's Generalized Binominal Theorem.

 

[robert Ihnot]

Vid: There's also a $10 dover book for those really interested.

Thanks! That is certainly interesting. I can remember thinking about those things, but I did not recognize there was so much written on it.

 

[Kurret]

My proof would look like this, which i think is easy to understand:
first the proof for y=x^2
$$f(x)=x^2$$
$$f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{(x+h)^2- x^2}{h}=\lim_{\substack{h\rightarrow 0}}\frac{2xh+h^2}{h}=\lim_{\substack{h\rightarrow 0}}2x+h=2x$$
now let $$f(x)=x^{\frac{1}{2}}$$
equivalent to:
$$f(x)^2=x$$
differentiate both side with respect to x, and using chainrule and the statement for x^2 we proved earlier:
$$2f(x)\cdot f'(x)=1$$
$$f'(x)=\frac{1}{2f(x)}=\frac{1}{2x^{\frac{1}{2}}}$$
this proof can easily be generalized to all rational numbers

 

[Big-T]

Kurret: This is not the concept discussed, you have taken the first derivative of some function. The topic is fractional derivatives (for example: what does it mean to take the pi'th derivative of a function?) not the derivatives of functions of fractional powers.,

 

[Kurret]

Kurret: This is not the concept discussed, you have taken the first derivative of some function. The topic is fractional derivatives (for example: what does it mean to take the pi'th derivative of a function?) not the derivatives of functions of fractional powers.,

Yes the discussion may have changed to that, but the thread starters question was about the first derivative of x^(1/2).

 

[Big-T]

I agree, sorry.

 

[gamesguru]

This is easy, and the way I do it can be applied to any rational exponent.
$$y=\sqrt{x}$$
$$y^2=x$$
$$2y \frac{dy}{dx}=1$$
$$\frac{dy}{dx}=\frac{1}{2y}=\frac{1}{2\sqrt{x}}$$.