Can someone show me the steps to finding the derivative of a cube root?

Question

agentnao · Answer

@inkyvoyd can you help me with this by any chance?

phi · Answer

change it to the 1/3 power
and use the power rule

zepdrix · Answer

So the trick issssss,
You want to rewrite the root as a fractional exponent. This will make it easier to differentiate.
Examples:
$$\large \sqrt{x}=x^{1/2}$$$$\large \sqrt[4]{x}=x^{1/4}$$
The degree of the root goes in the denominator of the fraction! :O
From there, you apply the power rule. are you familiar with the power rule? :)

agentnao · Answer

So then $$\frac{ 1 }{ 3 }x ^{-1/3}$$ ?

agentnao · Answer

Oops, I meant -2/3 as the exponent!

agentnao · Answer

OH! Wow. Yes, I see! I get it now :D

zepdrix · Answer

Ah yes good job! :D

agentnao · Answer

Thank you so you much! :D In fact, I have another question for you if you don't mind.

agentnao · Answer

$$x ^{2}-xy+y ^{2} = 1$$
They want the derivative of this as well. Would I insert $$\frac{ d _{y} }{ d _{x} }$$ next to the derivatives of both variables?

phi · Answer

the derivative with respect to x?
yes do it term by term

d/dx x^2  is 2 x dx/dx  or just 2x

xy will need the product rule

phi · Answer

the product rule is
d (u v) = u dv + v du

zepdrix · Answer

Since you're differentiating WITH RESPECT TO X, any time you take the derivative of something involving X, you have no problems.

But whenever you differentiate ANOTHER variable, like y, you'll have a dy/dx pop out.

So, you'll only multiply by dy/dx when you're differentiating a Y term.

Example:
$$\large (d/dx)x^2=2x$$
$$\large (d/dx)y^2=2y \frac{ dy }{ dx }$$

zepdrix · Answer

Phi's explanation was prolly clear enough, I just wanted to give an example in case you needed to see it :D

agentnao · Answer

Adding the y example answered the question I was just about to answer! Thank you Phi and Zepdrix! That explained it perfectly.

phi · Answer

I forgot to mention  
for d/dx x^2  you get 2x dx/dx   which is 2x
for d/dx y^2   you get 2y dy/dx   here you keep the dy/dx

btw, if you were doing 
d/dy  y^2  you would get just 2y
and
d/dy x^2  is 2x dx/dy  (don't often see this, but you might...)

phi · Answer

your final answer for
$$ \frac{d}{dx} (x^2 -xy +y^2-1 =0)$$ 
should be
$$ 2x - x \frac{dy}{dx} - y + 2y \frac{dy}{dx} =0 $$

you can solve for dy/dx   This is an example of implicit differentiation

agentnao · Answer

That's exactly what we were doing in class today! Okay, that makes sense. Thank you!