1. Implicit differentiation

Need help to finish this question,also how do I post in TeXnicCenter?

Find dy/dx and d^2y/dx^2 for x^2y - sin(x^2 - y) = 20

2x. dy/dx - cos(x^2 - y) (2x - dy/dx) = 0

2x. dy/dx = cos(x^2 - y) (2x - dy/dx)

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

dy/dx . (dx/2x - dy) = cos(x^2 - y)/2x

can't factor out dy/dx

??????????

Thanks.

$\displaystyle x^2y - \sin(x^2 - y) = 20$

where y is a function of x.

$\displaystyle x^2y'+2xy-\cos(x^2-y) \times (2x-y')=0$

Can you solve for y'?

3. Originally Posted by ashura
Need help to finish this question,also how do I post in TeXnicCenter?

Find dy/dx and d^2y/dx^2 for x^2y - sin(x^2 - y) = 20

2x. dy/dx - cos(x^2 - y) (2x - dy/dx) = 0

2x. dy/dx = cos(x^2 - y) (2x - dy/dx)

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

dy/dx . (dx/2x - dy) = cos(x^2 - y)/2x

can't factor out dy/dx

??????????

Thanks.
If you are asking how to post in LaTeX, see the first post in the thread here.

There seems to be a common problem in your implicit differentiation. For example, let's take your first term:
$\displaystyle x^2y$

There are two terms here, so you need to use the product rule, not just take the derivative of both. ie:
$\displaystyle \frac{d}{dx} \left ( f(x) \cdot g(x) \right ) = f'(x)g(x) + f(x)g'(x)$
not $\displaystyle f'(x)g'(x)$.

So
$\displaystyle \frac{d}{dx}(x^2y) = 2x \cdot y + x^2y'$

$\displaystyle x^2y - sin(x^2 - y) = 20$

Taking the first derivative of both sides:
$\displaystyle 2xy + x^2y' - cos(x^2-y) \cdot (2x-y') = 0$

$\displaystyle 2xy + x^2y' - (2x-y')cos(x^2-y) = 0$

$\displaystyle x^2y' +y'cos(x^2-y) = -2xy +2xcos(x^2-y)$

$\displaystyle y' [x^2 +cos(x^2-y)] = -2xy +2xcos(x^2-y)$

$\displaystyle y' = \frac{-2xy +2xcos(x^2-y)}{x^2 +cos(x^2-y)}$

To get y'' it would probably be easier to pick one of the line above before the division. (To get around that pesky quotient rule.)

$\displaystyle 2xy + x^2y' - (2x-y')cos(x^2-y) = 0$

Take the next derivative:
$\displaystyle [2y + 2xy'] + [2xy' + x^2y''] -$ $\displaystyle [(2-y'')cos(x^2-y) + (2x - y') \cdot -sin(x^2-y) \cdot (2x-y')] = 0$
(I have put [] around derivatives of individual terms above.)

You can solve for y'' from here. It will get a bit messy at some point because you should replace the y' in your answer for y'' with the expression you first got above. (In other words sub in your solution for y' in your final answer so that you have y'' as a function of just x and y.)

-Dan

4. Wow. A lot more texing that I wanted to do. Bravo.

5. Originally Posted by Jameson
Wow. A lot more texing that I wanted to do. Bravo.

-Dan

6. Thanks, didn't see the product in x^2y. Below is what i've worked out. Appreciate suggestions.

$\displaystyle [2y + 2xy'] + [2xy'+ x^2y'']-$
$\displaystyle [(2-y'')cos(x^2-y) +(2x-y').-sin(x^2-y).(2x-y')] = 0$

$\displaystyle Let A = [2y + 2xy']$
$\displaystyle Let B = [2xy'+ x^2y'']$
$\displaystyle Let C = [(2-y'')cos(x^2-y) + (2x-y').-sin(x^2-y).(2x-y')]$

$\displaystyle Let E = (x^2-y)$
$\displaystyle Let F = (2x-y')$

$\displaystyle A + B - C = 0$
$\displaystyle B - C = - A$

$\displaystyle B - C =:$
$\displaystyle [2xy'+ x^2y'']- [(2-y'')cos(E) - Sin(E) \times F^2]$

$\displaystyle [[x^2y'' + y''cos(E)]] + 2xy'- 2.cos(E)- Sin(E) \times F^2$

$\displaystyle Let D = 2xy'- 2.cos(E)- Sin(E) \times F^2$

$\displaystyle x^2y'' + y''cos(E) + D = - A$

$\displaystyle x^2y'' + y''cos(E) = -A - D$

$\displaystyle y''(x^2 + cos(E))= - A - D$

$\displaystyle y'' = \displaystyle\frac{-A - D}{(x^2 + cos(E)})$

$\displaystyle y'' = \displaystyle\frac{-(A + D)}{(x^2 + cos(E))}$

will substitute for A,B,C,D,E and F.

7. I don't feel like changing all the code for you, BUT no opening tags are needed like documentclass, etc. the  code should be replaced by [ math ] [/ math ] without the space.