Question

Using implicit differentiation, differentiate the following relation:

$\boxed{\huge\text{ \displaystyle g\left(x, y\right)=\left(3y^2+2y+6\right)x+y=0 }}$

And hence, at point (0, 0), the derivative is:
$\Large\text{ \displaystyle\frac{dy}{dx}\,g(x,y) = ... }$

Answer 1

Answer:

  $\dfrac{dy}{dx}g(0,0)=-6$

Step-by-step explanation:

You want the derivative dy/dx of the relation g(x, y) = 0 at the point (0, 0) where g(x, y) = x(3y² +2y+6) +y.

Implicit differentiation

The derivative relation will be the sum of the partial derivatives with respect to x and the partial derivatives with respect to y:

  x'(3y² +2y +6) +(x(6y +2) +1)y' = 0

Then dy/dx is the ratio y'/x':

  $\dfrac{y'}{x'}=-\dfrac{3y^2+2y+6}{6xy+2x+1}$

When x = y = 0, this reduces to ...

  $\dfrac{dy}{dx}=-\dfrac{6}{1}\\\\\\\boxed{\dfrac{dy}{dx}=-6\qquad\text{at (x, y)=(0, 0) }}$

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Additional comment

We can solve the relation for x, then take the derivative:

  x = -y/(3y² +2y +6)

  dx/dy = (-(3y²+2y +6) +y(6y +2))/(3y² +2y +6)²

  dx/dy = (3y² -6)/(3y² +2y +6)²

At x = y = 0, this is -6/36 = -1/6, so dy/dx = 1/(-1/6) = -6.