How do you find the second derivative of #4x^2 +9y^2 = 36#?

Question

Aurora Alvarado · Accepted Answer

#(d^2y)/(dx^2)=-16/(9y^3)# differentiate all terms on both sides #color(blue)"implicitly with respect to x"# #rArr8x+18y.dy/dx=0# #rArrdy/dx=(-8x)/(18y)=-(4x)/(9y)# #"to obtain "(d^2y)/(dx^2)" differentiate " dy/dx" using the "color(blue)" quotient rule"# #(d^2y)/(dx^2)=(9y.(-4)-(-4x).(9dy/dx))/(81y^2)# #color(white)((d^2y)/(dx^2))=(-36y+36x.dy/dx)/(81y^2)# #color(white)((d^2y)/(dx^2))=(-36y+36x.(-(4x)/(9y)))/(81y^2)# #color(white)((d^2y)/(dx^2))=(-36y-(16x^2)/y)/(81y^2)# #color(white)((d^2y)/(dx^2))=(-36y^2-16x^2)/(81y^3)# #color(white)((d^2y)/(dx^2))=(-4(9y^2+4x^2))/(81y^3)# #"now "9y^2+4x^2=36larr" initial statement"# #rArr(d^2y)/(dx^2)=-144/(81y^3)# #color(white)(rArr(d^2y)/(dx^2))=-16/(9y^3)#

Isaiah Allen · Answer

undefined To find the second derivative of $4x^2 + 9y^2 = 36$, we need to find the first derivative with respect to $x$ implicitly, then differentiate again to find the second derivative.

First, we differentiate the equation $4x^2 + 9y^2 = 36$ with respect to $x$ using implicit differentiation:

$$ \frac{d}{dx}(4x^2) + \frac{d}{dx}(9y^2) = \frac{d}{dx}(36) $$

$$ 8x + 18y \frac{dy}{dx} = 0 $$

Then, solving for $\frac{dy}{dx}$, we get:

$$ \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y} $$

Now, to find the second derivative, we differentiate $\frac{dy}{dx}$ with respect to $x$:

$$ \frac{d}{dx}\left(-\frac{4x}{9y}\right) = -\frac{4}{9}\frac{d}{dx}\left(\frac{x}{y}\right) $$

Using the quotient rule where $u = x$ and $v = y$, we get:

$$ -\frac{4}{9}\frac{d}{dx}\left(\frac{x}{y}\right) = -\frac{4}{9}\left(\frac{1 \cdot y - x \cdot \frac{dy}{dx}}{y^2}\right) $$

Substituting $\frac{dy}{dx} = -\frac{4x}{9y}$, we get:

$$ -\frac{4}{9}\left(\frac{y + \frac{4x^2}{9y}}{y^2}\right) $$

$$ = -\frac{4}{9}\left(\frac{9y^2 + 4x^2}{9y^3}\right) $$

$$ = -\frac{4}{9}\left(\frac{9y^2 + 4x^2}{(3y)^3}\right) $$

$$ = -\frac{4(9y^2 + 4x^2)}{27y^3} $$