Using implicit differentiation to find the slope of the tangent line

Stuck on this problem..

http://i55.tinypic.com/xp36gm.png

If it were using explicit differentiation, I could do this, but implicitly I'm having some trouble.

What I tried was implicitly differentiating both sides:

(d/dx) [ (4x + 3y)^(1/2) + (2xy)^(1/2) ] = 0

and after differentiating, I ended up with with:

(1/2)*(4x + 3y) * (4+3y') + 2(y + xy') = 0

From there, I plugged in the x,y values (3,8) for x and y, then solved for y', getting..

18 * (4 + 3y') + 2(8 + 3y') = 0

60y' = -88

y = -88/60

which is incorrect.

Am I approaching the problem from the wrong angle?

Re: Using implicit differentiation to find the slope of the tangent line

Your square roots magically disappeared.

Still a square root in there!

Re: Using implicit differentiation to find the slope of the tangent line

Re: Using implicit differentiation to find the slope of the tangent line

Nevermind, please delete. :)

Re: Using implicit differentiation to find the slope of the tangent line

Quote:

Originally Posted by

**Chronoc2005** Stuck on this problem..

http://i55.tinypic.com/xp36gm.png
If it were using explicit differentiation, I could do this, but implicitly I'm having some trouble.

What I tried was implicitly differentiating both sides:

(d/dx) [ (4x + 3y)^(1/2) + (2xy)^(1/2) ] = 0

and after differentiating, I ended up with with:

(1/2)*(4x + 3y) * (4+3y') + 2(y + xy') = 0

From there, I plugged in the x,y values (3,8) for x and y, then solved for y', getting..

18 * (4 + 3y') + 2(8 + 3y') = 0

60y' = -88

y = -88/60

which is incorrect.

Am I approaching the problem from the wrong angle?

The first of these derivatives...

Let .

.

So

The second of these derivatives.

Let .

.

So

Therefore, going back to your original equation...