# Thread: Differentiate using implicit differentiation

1. ## Differentiate using implicit differentiation

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

Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.

Answer6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy

2. Originally Posted by math34
find dy/dx (3xy)^(1/2)=3x-2y

Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.

Answer6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy
$\displaystyle \sqrt{3xy} = 3x - 2y$

So
$\displaystyle \frac{1}{2} \frac{1}{\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'$

$\displaystyle \frac{3y}{2\sqrt{3xy}} + \frac{3x}{2\sqrt{3xy}}y' = 3 - 2y'$

$\displaystyle \frac{3x}{2\sqrt{3xy}}y' + 2y' = 3 - \frac{3y}{2\sqrt{3xy}}$

$\displaystyle \left ( \frac{3x}{2\sqrt{3xy}} + 2 \right )y' = 3 - \frac{3y}{2\sqrt{3xy}}$

$\displaystyle y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2}$

$\displaystyle y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2} \cdot \frac{2 \sqrt{3xy}}{2 \sqrt{3xy}}$

$\displaystyle y' = \frac{6\sqrt{3xy} - 3y}{3x + 4\sqrt{3xy}}$ <-- To get your given answer, multiply top and bottom by $\displaystyle \sqrt{3xy}$ again.

We should rationalize this, but it is easier to use the original equation:
$\displaystyle \sqrt{3xy} = 3x - 2y$

$\displaystyle y' = \frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}$

$\displaystyle y' = \frac{18x - 12y - 3y}{3x + 12x - 8y}$

$\displaystyle y' = \frac{18x - 15y}{15x - 8y}$

-Dan

3. Hello, math34!

Find $\displaystyle \frac{dy}{dx}:\;\;(3xy)^{\frac{1}{2}} \:=\:3x-2y$

Would it be easier to first square both sides to get rid of the fractional exponent,
then differentiate implicitly? . . . absolutely!

Answer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!

Square: .$\displaystyle 3xy \:=\:9x^2 - 12xy + 4y^2$

We have: .$\displaystyle 3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'$

Then: .$\displaystyle 15xy' - 8yy' \:=\:18x - 15y$

Factor: .$\displaystyle (15x - 8y)y' \:=\:18x - 15y$

Therefore: .$\displaystyle \boxed{y' \:=\:\frac{18x - 15y}{15x - 8y}}$ . . . which is Dan's answer.

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It can be done head-on . . .

Differentiate implicitly: .$\displaystyle \frac{1}{2}(3xy)^{-\frac{1}{2}}(3y + 3xy') \:=\:3 - 2y'$

We have: .$\displaystyle 3y + 3xy' \:=\:2\sqrt{3xy}(3 - 2y')$

We are told that: .$\displaystyle \sqrt{3xy} \:=\:3x - 2y$

. . Hence, we have: .$\displaystyle 3y + 3xy' \:=\:2(3x - 2y)(3 - 2y')$

Expand: .$\displaystyle 3y + 3xy' \:=\:18x - 12xy' - 12y + 8yy'$

Simplify: .$\displaystyle 15xy' - 8yy' \:=\:18x - 15y$

Factor: .$\displaystyle (15x - 8y)y' \:=\:18x - 15y$

Therefore: .$\displaystyle \boxed{y' \:=\:\frac{18x-15y}{15x-8y}}$

4. ## Thanks

I got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!

Thanks guys

5. I am not happy about the idea of squaring the equation, and here's why:

Say we have:
$\displaystyle y = x$

Square both sides:
$\displaystyle y^2 = x^2$

Now take the derivative:
$\displaystyle 2yy' = 2x$

$\displaystyle y' = \frac{x}{y}$

Now, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.

Just something to watch out for.

-Dan