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.

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.

$\sqrt{3xy} = 3x - 2y$

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

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

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

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

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

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

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

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

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

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

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

-Dan

3. Hello, math34!

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

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

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

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

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

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

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It can be done head-on . . .

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

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

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

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

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

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

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

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

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:
$y = x$

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

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

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

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