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
$\displaystyle {\color{blue}\sqrt{3x+2y}} + {\color{red}\sqrt{3xy}} = 6.75$
$\displaystyle {\color{blue}(3x+2y)^{\frac{1}{2}}} + {\color{red}(3xy)^{\frac{1}{2}}} = 6.75$
Differentiating:
$\displaystyle \underbrace{{\color{blue} \frac{1}{2}(3x+2y)^{-\frac{1}{2}} \cdot (3 + 2y')}}_{\mbox{Chain rule}} + {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} \cdot } \underbrace{{\color{red}3 (y + xy')}}_{\mbox{Product Rule}} = 0$
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.
To make it clearer, let:
$\displaystyle a = {\color{blue} \frac{1}{2}(3x + 2y)^{-\frac{1}{2}} }$
$\displaystyle b = {\color{red} \frac{1}{2}(3xy)^{-\frac{1}{2}} }$
So we have:
$\displaystyle a(3 + 2y') + 3b(y + xy') = 0$
$\displaystyle 3a + {\color{red}2ay'} + 3by + {\color{red}3bxy'}$
Notice how you can isolate y' now? You probably got confused underneath all that garbage going on eh?
Keeping the same equivalences for a and b (kind of lazy):
$\displaystyle a(3 + 2y') + 3b(y + xy') = 0$
$\displaystyle 3a + 2ay' + 3by + 3bxy' = 0$
$\displaystyle 2ay' + 3bxy' = -3a - 3by$
$\displaystyle y'(2a + 3bx) = -3(a + by)$
$\displaystyle y' = \frac{-3(a + by)}{2 a+ 3bx}$
Then plug in x = 1, y = 4.
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):
$\displaystyle y' = \frac{-3\left[(3x + 2y)^{-\frac{1}{2}} + y(3xy)^{-\frac{1}{2}}\right]}{2\left[(3x + 2y)^{-\frac{1}{2}} + \frac{3x}{2}(3xy)^{-\frac{1}{2}} \right]}$
Now, plugging in (1,4):
$\displaystyle y' = \frac{-3 \left[(3(1) + 2(4))^{-\frac{1}{2}} + (4)\left[3(1)(4)\right]^{-\frac{1}{2}} \right]}{2 \left[(3(1) + 2(4))^{-\frac{1}{2}} + \frac{3(1)}{2}(3(1)(4))^{-\frac{1}{2}} \right]}$
$\displaystyle y' = \frac{-3\left[(11)^{-\frac{1}{2}} + 4(12)^{-\frac{1}{2}} \right]} {2 \left[ (11)^{-\frac{1}{2}} + \frac{3}{2}(12)^{-\frac{1}{2}}\right]}$