Implicit Differentiation

**EricK** · April 12th 2008, 01:17 PM

Find the slope of the tangent line to the curve

points (1,4)

i can get it to the point where you need to isolate the derivative, but im not sure how its done in this example

**o_O** · April 12th 2008, 01:34 PM

${\color{blue}\sqrt{3x+2y}} + {\color{red}\sqrt{3xy}} = 6.75$
${\color{blue}(3x+2y)^{\frac{1}{2}}} + {\color{red}(3xy)^{\frac{1}{2}}} = 6.75$

Differentiating:

$\underbrace{{\color{blue} \frac{1}{2}(3x+2y)^{-\frac{1}{2}} \cdot (3 + 2y')}}_{\mbox{Chain rule}} + {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} \cdot } \underbrace{{\color{red}3 (y + xy')}}_{\mbox{Product Rule}} = 0$

Now you have to do a bit of algebra to isolate y' but it should be doable. Collect all the terms with y' on one side and everything else on the other. Factor out y' and you should be good. Then plug in (1,4) to your end result.

**EricK** · April 12th 2008, 01:36 PM

thats exactly the point im stuck at. How do you isolate y' when both of them are in products?

**o_O** · April 12th 2008, 01:46 PM

To make it clearer, let:
$a = {\color{blue} \frac{1}{2}(3x + 2y)^{-\frac{1}{2}} }$
$b = {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} }$

So we have:
$a(3 + 2y') + 3b(y + xy') = 0$
$3a + {\color{red}2ay'} + 3by + {\color{red}3bxy'}$

Notice how you can isolate y' now? You probably got confused underneath all that garbage going on eh?

**EricK** · April 12th 2008, 02:02 PM

I dont understand why a(3+2y') is 3a +2y' and not 3a +2ay'. The same for the latter example. Why is that so? You can just pull the y' out even if its part of a product?

shouldnt b= 1/2(3xy)^-1/2

**o_O** · April 12th 2008, 02:20 PM

Oh dear. I made so many typos.

Yes, I did mean a(3x + 2y) = 3ax + 2ay'.

Yes, I did mean b= 1/2(3xy)^-1/2.

Sorry about that. I hope you get the idea though.

**EricK** · April 12th 2008, 02:52 PM

can you go further and solve it? Ive tried numerous ways and ive asked multiple people and no1 can get the answer

**o_O** · April 12th 2008, 07:23 PM

Keeping the same equivalences for a and b (kind of lazy

):
$a(3 + 2y') + 3b(y + xy') = 0$
$3a + 2ay' + 3by + 3bxy' = 0$
$2ay' + 3bxy' = -3a - 3by$
$y'(2a + 3bx) = -3(a + by)$
$y' = \frac{-3(a + by)}{2 a+ 3bx}$

Then plug in x = 1, y = 4.

**EricK** · April 12th 2008, 07:26 PM

once again though what happened to 2ay' and 3bxy'?

**o_O** · April 12th 2008, 07:32 PM

Ah sorry, guess reading through the latex causes me to miss little things. anyway, I edited my post for ya

**EricK** · April 12th 2008, 07:46 PM

there has to be something wrong with that unless im doing some little arithmatic error wrong. I'm doing this on a webwork system, so it tells me when its wrong.

**o_O** · April 12th 2008, 07:52 PM

Most likely. It's pretty messy to plug in. What did you get for your answer?

**EricK** · April 12th 2008, 07:59 PM

I've had so many answers... lets just say for arguements sake .4017475548, which is no way the answer.
Ive had answers like that, negative answers, answers like 6.2, 5.45, etc

**o_O** · April 12th 2008, 08:15 PM

Well it should be negative. Ok well, I wanted to avoid typing it all out but ...

I started from scratch and found y' and even did wrote it all neatly on paper to make sure I didn't make a mistake (which I hopefully didn't):
$y' = \frac{-3\left[(3x + 2y)^{-\frac{1}{2}} + y(3xy)^{-\frac{1}{2}}\right]}{2\left[(3x + 2y)^{-\frac{1}{2}} + \frac{3x}{2}(3xy)^{-\frac{1}{2}} \right]}$

Now, plugging in (1,4):

$y' = \frac{-3 \left[(3(1) + 2(4))^{-\frac{1}{2}} + (4)\left[3(1)(4)\right]^{-\frac{1}{2}} \right]}{2 \left[(3(1) + 2(4))^{-\frac{1}{2}} + \frac{3(1)}{2}(3(1)(4))^{-\frac{1}{2}} \right]}$

$y' = \frac{-3\left[(11)^{-\frac{1}{2}} + 4(12)^{-\frac{1}{2}} \right]} {2 \left[ (11)^{-\frac{1}{2}} + \frac{3}{2}(12)^{-\frac{1}{2}}\right]}$

**EricK** · April 12th 2008, 08:23 PM

well i plugged it into my calc and i got -4.2166.....
answer was wrong though

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