Derivative proof with a fractional exponent

[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.
$$y=\sqrt{x}$$
$$y^2=x$$
$$2y \frac{dy}{dx}=1$$
$$\frac{dy}{dx}=\frac{1}{2y}=\frac{1}{2\sqrt{x}}$$.