Question

using the definition. find the derivative of g(x) = cube root of f(x)

Answer 1

u can use the chain rule's :
g ' = f' / 3*cuberoot [f^2]

Answer 2

no can't use chain rule

Answer 3

f(x) = .... ???

Answer 4

g(x)= \[\sqrt[3]{f(x)}\]

Answer 5

hmmm .... do u have the specific value of f(x) ? like f(x) = x+2 or what ...

Answer 6

nope. just f(x)

Answer 7

I must say I find this question curious - including the apparent constraints. It sounds like a 'from first principles' approach is required. Now whether you do the quotient and let some h-> 0, or proceed with epsilon/delta arguments then you are going to have to - implicitly or explicitly - assume or prove the chain rule. No amount of clever-wingspan notation will escape that aspect. So to be more exact : what is 'the definition' being referred to here ?

Answer 8

'clever-wingspan' is not what I wrote :-) :-)

Answer 9

Some sort of substitution might be necessary. You have
\[g'(x)=\lim_{h\to 0}\frac{\sqrt[3]{f(x+h)}-\sqrt[3]{f(x)}}{h}\]
If you can think of one that makes the numerator into a difference of cubes, I think you can get somewhere. Maybe something like
\[a=\sqrt[3]{f(x+h)}\\
b=\sqrt[3]{f(x)}\]

Answer 10

\[\lim_{h\to 0}\frac{a-b}{h}\cdot\frac{a^2+ab+b^2}{a^2+ab+b^2}\\
\lim_{h\to 0}\frac{(a-b)\left(a^2+ab+b^2\right)}{h\left(a^2+ab+b^2\right)}\\
\lim_{h\to 0}\frac{a^3-b^3}{h\left(a^2+ab+b^2\right)}\\
\lim_{h\to 0}\frac{\left(\sqrt[3]{f(x+h)}\right)^3-\left(\sqrt[3]{f(x)}\right)^3}{h\left[\left(\sqrt[3]{f(x+h)}\right)^2+\left(\sqrt[3]{f(x+h)}\right)\left(\sqrt[3]{f(x)}\right)+\left(\sqrt[3]{f(x)}\right)^2\right]}\\
\lim_{h\to 0}\frac{f(x+h)-f(x)}{h\left[\left(f(x+h)\right)^{2/3}+\left(f(x+h)f(x)\right)^{1/3}+\left(f(x)\right)^{2/3}\right]}\\
f'(x)\cdot\lim_{h\to 0}\frac{1}{\left[\left(f(x+h)\right)^{2/3}+\left(f(x+h)f(x)\right)^{1/3}+\left(f(x)\right)^{2/3}\right]}\]
I hope it's clear what I did in that last step. From here, it all works out nicely to
\[\frac{1}{3}\left[f(x)\right]^{-2/3}\cdot f'(x)\]