y = x^3 then x = y^1/3

Question

y = x^3 then x = y^1/3

0

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If y = x³ the x = y^1/3

Find the derivative of dy/dx and dx/dy, and henc show that dy/dx * dx/dy = 1.

My derivatives were: 3x² and 1/ (3 root3of(y²))

When I multiplied them together I got: x² / root3of(y²)). Where I got stuck.

Can you show me the correct answer?

Thanks.

Guest Jul 26, 2017

edited by Guest Jul 26, 2017

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#1

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As follows:

\(y=x^3\quad \frac{dy}{dx}=3x^2\\x=y^{1/3}\quad \frac{dx}{dy}=\frac{1}{3}y^{-2/3}\rightarrow \frac{1}{3y^{2/3}} \rightarrow \frac{1}{3(y^{1/3})^2}\rightarrow \frac{1}{3x^2}\)

Now you should be able to see that dy/dx*dx/dy = 1

.

Alan Jul 26, 2017

#2

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x+y

Guest Jul 27, 2017

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Alan · Answer 1 · 2017-07-26T10:24:51

As follows:

\(y=x^3\quad \frac{dy}{dx}=3x^2\\x=y^{1/3}\quad \frac{dx}{dy}=\frac{1}{3}y^{-2/3}\rightarrow \frac{1}{3y^{2/3}} \rightarrow \frac{1}{3(y^{1/3})^2}\rightarrow \frac{1}{3x^2}\)

Now you should be able to see that dy/dx*dx/dy = 1

.

y = x^3 then x = y^1/3

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y = x^3 then x = y^1/3

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