Taking the second derivative - implicit differentiation

The original equation is this:

2x^2 -3y^2 = 0

(2x^2 -3y^2)' = (0)'

4x - 6y(dy/dx) = 0

I got the first derivative....which is correct according to my solutions manual: (2x/3y)

I took the derivative of this:

4x - 6y(dy/dx) = 0

(4x - 6y(dy/dx))' = (0)'

and got this using the product rule:

(6y)' ..... 6(dy/dx)

(dy/dx)....(dy/dx)^2

4 - 6(dy/dx)*(dy/dx) - 6y*(dy/dx)^2 = 0 ----> correct according the solutions manual

__________________________________________________ __________________

**This is where I went wrong....**

**I tried to isolate (dy/dx)^2 here:**

4 = 6(dy/dx)^2 + 6y(dy/dx)^2

4 = [(dy/dx)^2][(6)(1+y)]

dividing out [(6)(1+y)]

2/3(1+y)=(dy/dx)^2

**This is how the solutions manual does it:**

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

=2(3y^2-2x^2)/9y^3

**= - (8/9y^3)**

What did I do wrong when trying to isolate (dy/dx)^2 ???

Also ... a side question... is there an easier way to type out this stuff? Like a program I can use?

Thank you to whoever answers this....I have been trying to figure this out for like 2 hours

1 Attachment(s)

Re: Taking the second derivative - implicit differentiation

I haven't done implicit differentiation in a while, so I might be missing something ... but I managed to get down to the next-to-last line in your book's solution. How they managed to simplify that final numerator to -8, that I don't know.

See the attached pdf.

A suggestion. Maybe your notation is confusing you. Remember that, though the first derivative can written dy/dx, the second derivative is NOT "(dy/dx)^2" and is not written that way. I suggest you write y' for the first derivative and y'' for the second (so you have to remember that you're differentiating with respect to x).

In the pdf, since I'd figured out y', I substituted for that in the equation for y''. I'm not sure I follow your presentation, so I don't know whether you did that or not.

I hope this helps.

Re: Taking the second derivative - implicit differentiation

Ok I sort of get it more now that I realize I was writing it down wrong but I still can only make it to the second to last line....

Re: Taking the second derivative - implicit differentiation

Hi girl19! :)

Quote:

Originally Posted by

**girl19** The original equation is this:

2x^2 -3y^2 = 0

(2x^2 -3y^2)' = (0)'

4x - 6y(dy/dx) = 0

I got the first derivative....which is correct according to my solutions manual: (2x/3y)

I took the derivative of this:

4x - 6y(dy/dx) = 0

(4x - 6y(dy/dx))' = (0)'

and got this using the product rule:

(6y)' ..... 6(dy/dx)

(dy/dx)....(dy/dx)^2

That last is not quite correct.

It should be

$\displaystyle \left(\frac{dy}{dx}\right)' = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$

In particular:

$\displaystyle \frac{d^2y}{dx^2} \ne \left(\frac{dy}{dx}\right)^2$

Quote:

4 - 6(dy/dx)*(dy/dx) - 6y*(dy/dx)^2 = 0 ----> correct according the solutions manual

__________________________________________________ __________________

**This is where I went wrong....**

**I tried to isolate (dy/dx)^2 here:**

4 = 6(dy/dx)^2 + 6y(dy/dx)^2

So this should be:

$\displaystyle 4=6\left(\frac{dy}{dx}\right)^2 + 6y\left(\frac{d^2y}{dx^2}\right)$

Now you can substitute what you found earlier: $\displaystyle \frac{dy}{dx} = \frac{2x}{3y}$, and get:

$\displaystyle 4=6\left(\frac{2x}{3y}\right)^2 + 6y\left(\frac{d^2y}{dx^2}\right)$

Quote:

4 = [(dy/dx)^2][(6)(1+y)]

dividing out [(6)(1+y)]

2/3(1+y)=(dy/dx)^2

**This is how the solutions manual does it:**

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

=2(3y^2-2x^2)/9y^3

**= - (8/9y^3)**

What did I do wrong when trying to isolate (dy/dx)^2 ???

So you mixed up $\displaystyle \frac{d^2y}{dx^2}$ and $\displaystyle \left(\frac{dy}{dx}\right)^2$.

Quote:

Also ... a side question... is there an easier way to type out this stuff? Like a program I can use?

Thank you to whoever answers this....I have been trying to figure this out for like 2 hours

Yes there is.

Not a program, but a special way of typing that works on this forum.

If you type for instance [TEX]x_1^2 + 2y_2 = 3[/TEX], you get:

$\displaystyle x_1^2 + 2y_2 = 3$

This is called $\displaystyle \LaTeX$.

1 Attachment(s)

Re: Taking the second derivative - implicit differentiation

I think it can be done in a simple manner. Please look at the attachment.

Attachment 27625