How do you find the second derivative of #x^2+y^2=1#?

Question

Emma Aldridge · Accepted Answer

#(d^2y)/(dx^2)=-1/y-x^2/y^3#

We use implicit differentiation as follows:

differentiating #x^2+y^2=1#, we get

#2x+2y(dy)/(dx)=0# i.e. #(dy)/(dx)=-x/y#

and differentiating it further

#2+2[(dy)/(dx)(dy)/(dx)+y*(d^2y)/(dx^2)}=0#

or #2+2[(dy)/(dx)]^2+2y(d^2y)/(dx^2)=0#

or #2+2x^2/y^2+2y(d^2y)/(dx^2)=0#

or #2y(d^2y)/(dx^2)=-2-2x^2/y^2#

or #(d^2y)/(dx^2)=-2/(2y)-2x^2/y^2*1/(2y)#

= #-1/y-x^2/y^3#

Charles Anctil · Answer

undefined To find the second derivative of $x^2 + y^2 = 1$, follow these steps:

1. Differentiate the equation implicitly with respect to $x$.
$$2x + 2y \frac{dy}{dx} = 0$$
2. Solve for $\frac{dy}{dx}$ to get:
$$\frac{dy}{dx} = -\frac{x}{y}$$
3. Differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right)$$
4. Use the quotient rule to differentiate:
$$\frac{d^2y}{dx^2} = -\frac{y\frac{d}{dx}(x) - x\frac{d}{dx}(y)}{y^2}$$
5. Differentiate $x$ and $y$ with respect to $x$ to get:
$$\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2}$$
6. Substitute $\frac{dy}{dx} = -\frac{x}{y}$ into the equation:
$$\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2}$$
7. Simplify to get the second derivative:
$$\frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2}$$