Derivative problem: for x^3 + xy - 2x = 1, find dy/dx by

[hank]

Here's the problem:

(a) For x^3 + xy - 2x = 1, find dy/dx by differentiating implicitly.
(b) Solve x^3 + xy - 2x = 1 for y as a function of x, and find dy/dx from the equation.
(c) Confirm that the two results are consistent by expressing the derivative in part (a) as a function of x alone.

Here's what I get for a:

x^3 + xy - 2x = 1

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

Here's what I get for b:

y = (1 + 2x - x^3) / x

dy/dx = (x^3 - 3x^2 - 1) / x^2

I have no idea what to do for (c).

Can someone tell me what I'm supposed to do and confirm my (a) and (b) are correct?

Best Regards,

--Hank Stalica

 

[stapel]

a) I agree with your answer, but it is generally best to show your work:

. . . . .x³ + xy - 2x = 1

. . . . .3x² + y + x(dy/dx) - 2 = 0

. . . . .x(dy/dx) = 2 - 3x² - y

. . . . .dy/dx = (2 - 3x² - y) / x

b) I agree with your expression for "y=", but how did you get your derivative? Please show your work for this (showing how you applied the Quotient Rule), as I do not get anything close to this.

c) You solved for "y=" in (b). Plug this in for "y" in your "dy/dx =" for (a). Simplify to show that the two "dy/dx" expressions are actually the same thing.

. . . . .dy/dx = (2 - 3x² - y) / x

. . . . . . . . . .= (2 - 3x² - [(1 + 2x - x³)/x]) / x

. . . . . . . . . .= (2 - 3x² - 1 - 2x + x³) / x²

...and so forth.

Eliz.

 

[hank]

Thanks for the reply.

Here's how I found the derivative of:
y = (1 + 2x - x^3) / x

Answer:
dy/dx = [x(2 - 3x^2) - (1 + 2x - x^3)(1)] / x^2
= (2x - 3x^2 - 1 - 2x + x^3) / x^2
= (x^3 - 3x^2 - 1) / x^2

 

[stapel]

dy/dx = [x(2 - 3x^2) - (1 + 2x - x^3)(1)] / x^2
= (2x - 3x^2 - 1 - 2x + x^3) / x^2

I think this should be:

. . . . .(2x - 3x³ - 1 - 2x + x³) / x²

I could be wrong, of course....

Eliz.