implicit differentiation of y=(cbrt 2x)

[calphobe^2]

y=(cbrt 2x)

d/dy (y) = d/dx (cbrt2x)

1) d/dy (y) = d/dx (2x^1/3) ?
2)

Please,What are the steps to chain rule this?
Thanks

 

[galactus]

Is this what you're trying to differentiate?:

\(\displaystyle \L\\y=\sqrt[3]{2x}\)

This seems rather straightforward. What's the problem?.

 

[calphobe^2]

Yes. That is what I'm trying to differentiate.
I'm getting:
y= cbrt(2x)

=2x^(1/3)


dy/dx= ( 2/3 ) (x^ (-2/3))

= 2/3x^(2/3)

I'm missing something, and know this is very basic. Please pardon my math infantilism-I'm wanting to overcome it. The result sought is:

2^(1/3) / 3x^(2/3) ...Where does the (1/3) power in the numerator come from?

 

[galactus]

That's OK. Let's see.

\(\displaystyle \L\\\frac{d}{dx}[(2x)^{\frac{1}{3}}]\\)

=\(\displaystyle \L\\\frac{1}{3}(2x)^{\frac{-2}{3}}\underbrace{(2)}_{\text{chain rule applied}}\)

=\(\displaystyle \L\\\frac{2}{3(2x)^{\frac{2}{3}}}\)

Multiply numerator and denominator by \(\displaystyle \L\\2^{\frac{-2}{3}}\)

That gives you your required form of \(\displaystyle \L\\\frac{2^{\frac{1}{3}}}{3x^{\frac{2}{3}}\)

Another way to get that form is to use the product rule.

\(\displaystyle 2^{\frac{1}{3}}x^{\frac{1}{3}}=2^{\frac{1}{3}}(\frac{1}{3}x^{\frac{-2}{3}})=\frac{2^{\frac{1}{3}}}{3x^{\frac{2}{3}}\)

Frankly, I'd stick with \(\displaystyle \frac{2}{3(2x)^{\frac{2}{3}}}\)

 

[calphobe^2]

Thank you.