How do you find y'' by implicit differentiation for #4x^2 + 3y^2 = 6#?

Question

Luke Alvarez · Accepted Answer

#4x^2 + 3y^2 = 6# #d/(dx)(4x^2 + 3y^2) = d/(dx)(6)# #8x + 6y(dy)/(dx) = 0# So  #(dy)/(dx) = (-4x)/(3y)# On to the second derivative: #d/(dx)((dy)/(dx)) =d/(dx) ( (-4x)/(3y))# #(d^2y)/(dx^2) = ((-4)(3y)- (-4x)(3(dy)/dx))/((3y)^2)# #(d^2y)/(dx^2) = (-12y +12 x(dy)/dx)/(9y^2)#   Replacing #dy/dx# by the expression above gives us: #(d^2y)/(dx^2) = (-12y +12 x((-4x)/(3y)))/(9y^2) =  (-12y + (-16x^2)/y)/(9y^2)#   No multiply by #y/y# to get  #(d^2y)/(dx^2) = (-12y^2 -16x^2)/(9y^3) = (-4(4x^2+3y^2))/(9y^3)# Now, use the initial equation:  #4x^2 + 3y^2 = 6#  to write the answer as: #(d^2y)/(dx^2) = = (-24)/(9y^3) = (-8)/(3y^3)#

Paisley Johnson · Answer

undefined To find $ y'' $ by implicit differentiation for the equation $ 4x^2 + 3y^2 = 6 $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Apply the chain rule when differentiating $ y^2 $.
3. Solve for $ y'' $ in terms of $ x $ and $ y $.

First differentiation:
$$ \frac{d}{dx}(4x^2) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(6) $$

Simplify:
$$ 8x + 6y \frac{dy}{dx} = 0 $$

Now differentiate $ 6y \frac{dy}{dx} $ using the product rule:
$$ 6 \frac{dy}{dx} + 6y \frac{d^2y}{dx^2} = 0 $$

Isolate $ \frac{d^2y}{dx^2} $:
$$ 6y \frac{d^2y}{dx^2} = -6 \frac{dy}{dx} $$

$$ \frac{d^2y}{dx^2} = -\frac{1}{y} \frac{dy}{dx} $$

Substitute the value of $ \frac{dy}{dx} $ from the first derivative:
$$ \frac{d^2y}{dx^2} = -\frac{1}{6y} (8x) $$

$$ \frac{d^2y}{dx^2} = -\frac{4x}{3y} $$