# Math Help - Implicit differentiation

1. ## Implicit differentiation

1.x^2+3xy+y^2=15 And the answer says -y^2/x^2

For 1., dy/dx(x^2+3xy+y^2)=dy/dx(15)->(2x+3x+3y(dy/dx)+2y(dy/dx)=0 (for 3xy, I used the product rule for x and y and got x+y and plugged it to 3 to get this function.) What am i doing wrong with this equation?

With 2. I used the quotient rule, and got (x/x^2)+(y/y^2) and I know I am wrong on this one from the start. please help.

2. For #1:

$x^{2}+3xy+y^{2}=15$

$2x+3x\frac{dy}{dx}+3y+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-(3y+2x)}{3x+2y}$

Are you sure that's the book answer?. It isn't $\frac{-y^{2}}{x^{2}}$

3. Originally Posted by driver327
1.x^2+3xy+y^2=15 And the answer says -y^2/x^2

For 1., dy/dx(x^2+3xy+y^2)=dy/dx(15)->(2x+3x+3y(dy/dx)+2y(dy/dx)=0 (for 3xy, I used the product rule for x and y and got x+y and plugged it to 3 to get this function.) What am i doing wrong with this equation?

With 2. I used the quotient rule, and got (x/x^2)+(y/y^2) and I know I am wrong on this one from the start. please help.
Hello,

could it be that the first answer is missing?

To derivate implicitely I use the formula:

If $f(x,y) = 0$ then $\frac{df}{dx}=y'=-\frac{\frac{\partial{f}}{\partial{x}}}{\frac{\part ial{f}}{\partial{y}}}$

to #1:
$f(x,y) = x^2+3xy+y^2-15=0$

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

to #2:

Using the same method. $f(x,y) = \frac{1}{x} + \frac{1}{y} = \frac{x+y}{x\cdot y}$ then (using quotient rule!)

$y' = -\frac{\frac{xy - (x+y)y}{x^2 \cdot y^2}}{\frac{xy-(x+y)x}{x^2 \cdot y^2}} = -\frac{-y^2}{-x^2} = -\frac{y^2}{x^2}$

4. I accidentally put down the same answer. But the first one is correct.