Question

Implicitly differentiate sqrt(3x+2y)+sqrt(4xy)=21.1 and find the tangent line at (8,7).

Answer 1

Would I start like:
1/2(3x+2y)^-1/2 + 1/2(4xy)^-1/2 = 0

or:
(1/2(3x+2y)^-1/2)3x+2y + (1/2(4xy)^-1/2)4xy = 0

Answer 2

Also, since it's implicit differentiation, I need to put in the (dy/dx), but I'm not quite sure where it would fit.

Answer 3

d/dx(sqrt(3x+2y) = 1/2 (3x+2y)^-1/2 (3+ 2dy/dx)

was that understandable ?

Answer 4

I think so. So, then I differentiate to get
1/2(3x+2y)^-1/2 (3x+2(dy/dx)) + 1/2(4xy)^-1/2 (4x(dy/dx)) = 0

I need to get (dy/dx) by itself, so I can go:
1/2(3x+2y)^-1/2 (3x+2(dy/dx)) = -1/2(4xy)^-1/2 (4x(dy/dx))

...but then I'm stuck. If I divide either side by the term with y', I've still got one in the other side, but then it's in a nasty fraction. What am I doing wrong, here?

Answer 5

there is a product rule to be done on that second term

Answer 6

\[\frac{d}{dx} (4xy )= 4y + 4x \frac{dy}{dx}\]

Answer 7

Yes, I sure did. Thanks. Unless I'm missing the point, though, that still leaves me in more or less the same place.

(1/2(3x+2y)^-1/2)(3+2(dy/dx)) = (-1/2(4xy)^-1/2)(4y+4(dy/dx))

I still have a dy/dx as a factor on each side, and I don't know how to get it by itself. I feel like I'm making this harder than it needs to be.

Answer 8

get them both on the same side, then factor out dy/dx

Answer 9

\[\frac{3+2y'}{2\sqrt{3x+2y}}=\frac{4y+4xy'}{2\sqrt{4xy}}\]\[\frac3{2\sqrt{3x+2y}}+\frac{y'}{\sqrt{3x+2y}}=\frac {y}{\sqrt{xy}}+\frac{xy'}{\sqrt{xy}} \]can you see how to separate y' yet?

Answer 10

I think I've almost got it...

Answer 11

...or not.