Can I use L'Hospitals rule here? (seem like i use it too often )

[linuxux]

Does l'hospitals rule work here?:

$$\lim_{h\to_0}\frac{f\left(8+h\right)-f\left(8\right)}{h}$$ for $$f\left(x\right)=x^\frac{4}{3}$$

then i would get,

$$\lim_{h\to_0}\frac{f\left(8+h\right)}{1}$$

then,

$$=\frac{\left(8\right)^\frac{4}{3}}{1}$$ ?

is that it?...

 

[matt grime]

No, you can't use it there, in my opinion. The question is asking you to work out the derivative of x^{4/3} at x=8, thus you can't invoke l'Hopital which a priori assumes the function to be differentiable. A minor quibble, but given the way you've written it it seems clear that you're being asked to prove the derivative exists, so assuming it does is not allowed.

 

[linuxux]

i tried factoring according to this:

$$a^4-b^4=\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)$$

but I am stuck because i can't get a multiple of h to cancel the denomiator's h.

 

[matt grime]

Where did the power of 1/3 vanish to? (Note I don't have a simple solution in mind. But it is easy to show that x^4 is differentiable, as is x^3, and hence x^1/3 by the inverse function theorem, hence x^{4/3} is diffible, and it all follows some what simply from these big sounding theorems).

 

[linuxux]

...we haven't touched those theorems yet.

but, $$a=\left(8+h\right)^\frac{1}{3}$$ and $$b=\left(8\right)^\frac{1}{3}$$

 

[dextercioby]

x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n-1). Just plug n=4/3 and x=8 to see what you get.

 

[Gib Z]

x^n is differentiable at any "x", no matter the value of "n>0".

I think he's been asked to show that, at least for n= 4/3.

 

[linuxux]

x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n-1). Just plug n=4/3 and x=8 to see what you get.

I see it equals what i wrote using l'hospitals rule, but through what derivative what did you go through to determine that?