Derivative Problems

[mcb30]

We're going over basic differentiation, and the problem I had is to find the derivative of f(x)= x^(1/3). I have to use the definition of a derivative, but I'm not sure what to do when I begin with two cube roots in the numerator. Can anyone help?

[mezarashi]

I'm not sure what you mean by the definition of a derivative. Do you mean find the derivative using first principles? Or else you can use the power rule f(x)=x^a, where a=1/3.

[BobbyJoeCool]

[math]\frac{d}{dx}(ax^b)=(a*b)x^{b-1}[/math]

 

So, your problem:

 

[math]\frac{d}{dx}(x^{1/3})=(1*1/3)x^{(1/3)-1}[/math]

 

[math]\frac{x^{-2/3}}{3}=\frac{1}{3x^{2/3}}=\frac{1}{3\sqrt[3]{x^2}}[/math]

[oatjay]

i think he means to solve it using first principles- the whole fraction-y thing with the delta -x's

[BobbyJoeCool]

i think he means to solve it using first principles- the whole fraction-y thing with the delta -x's

 

ohh shudder..

 

you mean this way...

 

[math]f'(x)=\lim_{\Delta x\to 0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}[/math]

 

I call this "The evil way." Why... Subsitiute... this is what you get...

 

[math]f'(x)=\lim_{\Delta x\to 0}{\frac{\sqrt[3]{x+\Delta x}-\sqrt[3]{x}}{\Delta x}}[/math]

 

and it gets even more evil from there, because you have to get the Delta x out of the denominator so that you don't get 0 down there. And the way I would go about doing that is attempt to get a common factor of Delta x in the numerator that I could factor out that with the Delat x in the denominator... Oi vey. This isn't exactly the type of problem I would assign to do this way, especially in the beginning of a calculus class.

[oatjay]

i give up on that one, man. i tried messing around with it a bit, but i got very close to nowhere with it. algebra was never my strong point anyway. i guess this post is meaningless, but i'd be really really happy if someone figured this out and posted it because it's been annoying the hell out of me.

[BobbyJoeCool]

it's like... ridiculusly Advanced Algebra. How do you factor out the Delta x out of the denominator? The most obvious way is to factor a Delta X out each term of the numerator.

 

[math]\frac{\sqrt[3]{x- \Delta x}-\sqrt[3]{x}}{\Delta x}[/math]

 

good luck. the trick of multiplying by the congigate doesn't work, because you have cubed roots instead of square roots.

 

I tried this..

 

[math]\frac{(\sqrt[3]{x- \Delta x}-\sqrt[3]{x})(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})}{\Delta x(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})}[/math]

 

[math]\frac{(\sqrt[3]{x- \Delta x}^2-\sqrt[3]{x}^2)}{\Delta x(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})}[/math]

 

[math]\frac{(\sqrt[3]{x- \Delta x}^2-\sqrt[3]{x}^2)(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})}{\Delta x(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})(\sqrt[3]{x- \Delta x}+\sqrt[3]{x})}[/math]

 

for confusion factor, Delta x shall now be replaced by "y."

 

[math]\frac{x-y+\sqrt[3]{x-y}^2\sqrt[3]{x}-\sqrt[3]{x-y}\sqrt[3]{x}^2-x}{y}[/math]

 

[math]\frac{-y+\sqrt[3]{(x-y)^2}\sqrt[3]{x}-\sqrt[3]{x-y}\sqrt[3]{x^2}}{y}[/math]

 

[math]\frac{-y+\sqrt[3]{(x^2-2xy+y^2}\sqrt[3]{x}-\sqrt[3]{x-y}\sqrt[3]{x^2}}{y}[/math]

 

[math]\frac{-y+\sqrt[3]{x^3-2x^2 y+y^2 x}-\sqrt[3]{x^3-x^2 y}}{y}[/math]

 

I gave up here after staring at it for a while (a half hour). I realized that at an intro to calculus level, there's no way to finish solving this. And at any rate, it's not the type of problem you would think would be assigned if it's going to be this long, so you'd have to assume that they can use power rule to make it a simple math assignment problem. Either that or all of us are thinking too much. Where's (is it Matt Grime?) the Math Expert when I need him?

 

If it can be done this way, I'll figure it out eventually (or someone will tell me).

[oatjay]

yeah, i tried that too- i also tried to pull a delta x out of the numerator and canceled it with the denominator, but that didn't get me anywhere either. too many damn letters. aghhh.

[BobbyJoeCool]

the variables don't bother me, it's those freaking cubed roots... AHHHHHHH!!! blah blah blah...

[BobbyJoeCool]

ok. there's a formula for this...

 

[math]a^3-b^3=(a-b)(a^2+ab+b^2)[/math]

 

so lets look at the equation again!

 

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

 

[math]f'(x)=\frac{\sqrt[3]{x+ \Delta x}-\sqrt[3]{x}}{\Delta x}[/math]

 

Ok. again, for confusion, Delta x shall be replaced by y.

 

[math]\frac{\sqrt[3]{x+y}-\sqrt[3]{x}}{y}[/math]

 

[math]a=\sqrt[3]{x+y}[/math]

[math]b=\sqrt[3]{x}[/math]

 

Since the first term is already there (a-b) we multiply the num and denom by the second term.

 

[math]\frac{(\sqrt[3]{x+y}-\sqrt[3]{x})(\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2)} {y(\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2)}[/math]

 

by our original formula...

 

[math]x+y-x=(\sqrt[3]{x+y}-\sqrt[3]{x})(\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2)[/math]

 

[math]\frac{x+y-x} {y(\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2)}[/math]

 

[math]\frac{y} {y(\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2)}[/math]

 

the "y"s cancil out...

 

[math]\frac{1} {\sqrt[3]{x+y}^2+\sqrt[3]{x+y}\sqrt[3]{x}+\sqrt[3]{x}^2}[/math]

 

now remember y=Delta x

 

[math]\frac{1} {\sqrt[3]{x+\Delta x}^2+\sqrt[3]{x+\Delta x}\sqrt[3]{x}+\sqrt[3]{x}^2}[/math]

 

and it's a limit...

 

[math]\lim_{\Delta x \to 0}{\frac{1} {\sqrt[3]{x+\Delta x}^2+\sqrt[3]{x+\Delta x}\sqrt[3]{x}+\sqrt[3]{x}^2}}[/math]

 

So with direct substitution...

 

[math]{\frac{1} {\sqrt[3]{x+0}^2+\sqrt[3]{x+0}\sqrt[3]{x}+\sqrt[3]{x}^2}[/math]

 

[math]{\frac{1} {\sqrt[3]{x}^2+\sqrt[3]{x}\sqrt[3]{x}+\sqrt[3]{x}^2}[/math]

 

[math]{\frac{1} {2 \sqrt[3]{x}^2+\sqrt[3]{x}\sqrt[3]{x}}[/math]

 

[math]{\frac{1} {2 \sqrt[3]{x}^2+\sqrt[3]{x}^2}[/math]

 

[math]{\frac{1} {3 \sqrt[3]{x}^2}[/math]

 

[math]f'(x)={\frac{1} {3 \sqrt[3]{x}^2}[/math]

 

Which, by the power rule as I explained earlier, is the right answer.

 

Kudos to both me, and the Wartburg College Math department who spend all of 5 minutes explaining this to me (after spending 5 seconds figuring it out.)

[Dave]

One little nitpick, because I can't quite resist. You say:

 

[math]f'(x)=\frac{\sqrt[3]{x+ \Delta x}-\sqrt[3]{x}}{\Delta x}[/math]

 

Which isn't really true. f' is the limit of that fraction. Other than that, looks pretty good (although the squared on the cube root of x towards the end should be inside instead of outside).

[BobbyJoeCool]

One little nitpick' date=' because I can't quite resist. You say:

 

[math']f'(x)=\frac{\sqrt[3]{x+ \Delta x}-\sqrt[3]{x}}{\Delta x}[/math]

 

Which isn't really true. f' is the limit of that fraction. Other than that, looks pretty good

you are right of course... opps.

(although the squared on the cube root of x towards the end should be inside instead of outside).

I don't think it matters.... I mean... [math]\sqrt[3]{x}^2=\sqrt[3]{x^2}[/math]

[Dave]

Nah, just a matter of aesthetics on my part really

[BobbyJoeCool]

After all.. it's really should be written [math]x^{2/3}[/math]