<p>In class the other day, our teacher asked us to take the derivative of x^(1/3) or the cube root, using the definition. I'm in AP Calc BC and noone could figure it out. I remember a website that explained it, but does anyone know how to do it?</p>
<p>Hmm...it seems pretty simple to me and I'm just in AB. Wouldn't you just use the power rule to arrive at 1/3x^-2/3. Maybe I'm interpreting it the wrong way, but that seems like the answer.</p>
<p>using the definition not the power rules</p>
<p>[(x+h)^1/3 -x^1/3]/h, I'll work on it........</p>
<p>easier to do it by the other definition</p>
<p>got it
1</p>
<hr>
<p>[(x+h)^(2/3)] + [(x) ^(2/3)]</p>
<p>= 1/ (2x^(2/3))</p>
<p>ok let me explain, actually adidasty's setup is very good, all you have to do is rationalize the numerator by multiplying it by my denominator in the solution, then the h's cancel out easily</p>
<p>1/3 + 2/3 = 1</p>
<p>does anyone care to verify this?</p>
<p>Yeah, altf4....but you have 2x^(2/3), not 3 like the power rule says the answer should be.</p>
<p>ya, that's weird, hmmm</p>
<p>Looks like a hard algebra problem, not a calculus one.</p>
<p>i dont kno how i got them not to be the same...</p>
<p>Had this type of question on past test.
y=x^(1/3)
y^3=X
Take Derivative of each side.
3y^2 y'=1
y'=1/(3y^2)
=1/3[x^(1/3)^2]
=1/3x^(2/3)
=[1x^(-2/3)]/3</p>
<p>you're still using the power rule though</p>
<p>You know why it's so "weird"? Well, look at it this way:</p>
<p>STEP 1:
lim (h --> 0): [(x+h)^(1/3) - (x)^(1/3)] / h</p>
<p>and most of us are multiplying the above with </p>
<p>STEP 2:
<a href="(x+h)%5E2/3%20+%20(x)%5E2/3">u</a><a href="x+h">/u</a>^2/3 + (x)^2/3)</p>
<p>What happens is that you DON'T get a quotient such as </p>
<p>STEP 3:
h+x-x
h [(x+h)^(2/3)][(x)^2/3]</p>
<p>If you multiply steps 1 and 2 out with the foil method, you actually get something else:
[h + x - x + ((x)^2/3)((x+h)^1/3) - ((x)^1/3)((x+h)^2/3)]
h((x+h)^2/3)+((x)^2/3</p>
<p>simplified, you'd get:
[h + ((x)^2/3)((x+h)^1/3) - ((x)^1/3)((x+h)^2/3)]
h((x+h)^2/3)+((x)^2/3</p>
<p>So you see, it's not as "simple" as it first appears to be. I'll try to return with a follow-up (a solution, hopefully? haha) soon. Cheers!</p>
<p>let f(x) = x^(1/3)</p>
<p>Prove: f'(x) = (1/3)x^(-2/3) by the Power Rule</p>
<p>Proof:</p>
<p>By the definition of the derivative:</p>
<p>f'(x) = lim h-> 0 of</p>
<p>(x+h)^(1/3) - x^(1/3)</p>
<hr>
<pre><code> h
</code></pre>
<p>NOTE: for simplicity's sake, I'm just working with the algebra and leaving out the formal limit notation herein.</p>
<p>Now, recognizing the the difference of cubes we rationalize the numerator by multiplying it by its conjugate, shown below:</p>
<h2>(x+h)^(2/3)+(x+h)^(1/3)+x^(1/3)+x^(2/3)</h2>
<p>(x+h)^(2/3)+(x+h)^(1/3)+x^(1/3)+x^(2/3)</p>
<p>When multiplied out and simplified, we have:</p>
<pre><code> 1
</code></pre>
<hr>
<p>(x+h)^(2/3)+(x+h)^(1/3)+x^(1/3)+x^(2/3)</p>
<p>Taking the limit yields:</p>
<h2> 1</h2>
<p>x^(2/3) + x^(2/3) + x^(2/3)</p>
<p>Which simplifies to:
1</p>
<hr>
<pre><code> 3(x^(2/3))
</code></pre>
<p>Or:</p>
<p>f'(x) = (1/3)x^(-2/3)</p>
<p>I used this formula and arrived with the answer.....</p>
<p>f(x)=A<em>x^n then f ' (x) = n</em>A*x^(n-1) SO.......</p>
<p>f(x)=x^(1/3) f ' (x) 1/3<em>x^(1/3-1)= ANSWER----->1/(3</em>x^(2/3))<-----ANSWER</p>
<p>BUT WITH THE DEFINITION.....ITS HARD.....SO YOU KNOW THE ANSWER AT LEAST...;)</p>
<p>trog, please explain how to find the conjugate of a cubic root polynomial, slowly, i'm not following</p>
<p>marko, we already know the answer</p>
<h2>(x+h)^(2/3)+(x+h)^(1/3)+x^(1/3)+x^(2/3)</h2>
<p>(x+h)^(2/3)+(x+h)^(1/3)+x^(1/3)+x^(2/3)</p>
<p>why is it (x+h)^(1/3)+x^(1/3), as you stated above? the sum of cubes is X^2+XY+Y^2, so shouldn't the two elements be combined ((x+h)^(1/3)+x^(1/3))?</p>
<p>Yes, you're right altf4, I wrote it out wrong.</p>
<p>ok, thanks for clearing that up, now i get it, u multiply by whatever to complete the cube</p>