# Using implicit differentiation to find the slope of the tangent line

• Sep 9th 2011, 02:02 PM
Chronoc2005
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?
• Sep 9th 2011, 02:26 PM
TKHunny
Re: Using implicit differentiation to find the slope of the tangent line

$\frac{d}{dx}\sqrt{x}\;=\;\frac{1}{2\cdot\sqrt{x}}$

Still a square root in there!
• Sep 9th 2011, 04:07 PM
Chronoc2005
Re: Using implicit differentiation to find the slope of the tangent line
Aha, thank you much! :)
• Sep 11th 2011, 07:59 PM
Chronoc2005
Re: Using implicit differentiation to find the slope of the tangent line
• Sep 11th 2011, 09:26 PM
Prove It
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?

\displaystyle \begin{align*}\sqrt{4x + 3y} + \sqrt{2xy} &= 12.9 \\ \left(4x + 3y\right)^{\frac{1}{2}} + \left(2xy\right)^{\frac{1}{2}} &= 12.9 \\ \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}} + \left(2xy\right)^{\frac{1}{2}}\right] &= \frac{d}{dx}\left(12.9\right) \\ \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}}\right] + \frac{d}{dx}\left[\left(2xy\right)^{\frac{1}{2}}\right] &= 0 \end{align*}

The first of these derivatives...

$\displaystyle z = \left(4x + 3y\right)^{\frac{1}{2}}$

Let $\displaystyle u = 4x + 3y \implies z = u^{\frac{1}{2}}$.

$\displaystyle \frac{du}{dx} = 4 + 3\frac{dy}{dx}$.

\displaystyle \begin{align*} \frac{dz}{du} &= \frac{1}{2}u^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{4x + 3y}} \end{align*}

So $\displaystyle \frac{dz}{dx} = \frac{4 + 3\frac{dy}{dx}}{2\sqrt{4x + 3y}}$

The second of these derivatives.

$\displaystyle w = \left(2xy\right)^{\frac{1}{2}}$

Let $\displaystyle u = 2xy \implies w = u^{\frac{1}{2}}$.

$\displaystyle \frac{du}{dx} = 2y + 2x\,\frac{dy}{dx}$.

\displaystyle \begin{align*} \frac{dw}{du} &= \frac{1}{2}u^{-\frac{1}{2}} \\ &= \frac{1}{2\sqrt{2xy}} \end{align*}

So $\displaystyle \frac{dw}{dx} = \frac{2y + 2x\,\frac{dy}{dx}}{2\sqrt{2xy}}$

Therefore, going back to your original equation...

\displaystyle \begin{align*} \frac{d}{dx}\left[\left(4x + 3y\right)^{\frac{1}{2}}\right] + \frac{d}{dx}\left[\left(2xy\right)^{\frac{1}{2}}\right] &= 0 \\ \frac{4 + 3\frac{dy}{dx}}{2\sqrt{4x + 3y}} + \frac{2y + 2x\,\frac{dy}{dx}}{2\sqrt{2xy}} &= 0 \\ \frac{\left(4 + 3\frac{dy}{dx}\right)\sqrt{2xy} + \left(2y + 2x\,\frac{dy}{dx}\right)\sqrt{4x + 3y}}{2\sqrt{4x + 3y}\sqrt{2xy}} &= 0 \\ \left(4 + 3\frac{dy}{dx}\right)\sqrt{2xy} + \left(2y + 2x\,\frac{dy}{dx}\right)\sqrt{4x + 3y} &= 0 \\ 4\sqrt{2xy} + 3\sqrt{2xy}\,\frac{dy}{dx} + 2y\sqrt{4x + 3y} + 2x\sqrt{4x + 3y}\,\frac{dy}{dx} &= 0 \\ \left(3\sqrt{2xy} + 2x\sqrt{4x + 3y}\right)\frac{dy}{dx} &= -4\sqrt{2xy} - 2y\sqrt{4x + 3y} \\ \frac{dy}{dx} &= \frac{-2\left(2\sqrt{2xy} + y\sqrt{4x + 3y}\right)}{3\sqrt{2xy} + 2x\sqrt{4x + 3y}}\end{align*}