+0

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

0
265
2

If y = x3  the x = y1/3

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

#1
+26677
+3

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
Sort:

#1
+26677
+3

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
-1

x+y

Guest Jul 27, 2017