Question #ce2e0

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Question #ce2e0

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Answer 1 · 2016-11-12T16:18:33

Although we could solve for $y$ $y$ , this looks like a classic implicit differentiation exercise.

Explanation:

I assume that you want the second derivative of $y$ $y$ with respect to $x$ $x$ . (Yes, there are other possibilities.)

$x^{2} + y^{2} = 90$ $x^2+y^2=90$

Find $\frac{d y}{d x}$ $dy/dx$

$\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (90)$ $d/dx(x^2+y^2) = d/dx(90)$

$2 x + 2 y \frac{d y}{d x} = 0$ $2x+2y dy/dx = 0$ , so

$\frac{d y}{d x} = - \frac{x}{y}$ $dy/dx = -x/y$

Differentiate again:

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d^{2} y}{{d x}^{2}} = \frac{(- 1) (y) - (- x) (\frac{d y}{d x})}{y^{2}}$ $d/dx(dy/dx) = (d^2y)/dx^2 = ((-1)(y)-(-x)(dy/dx))/y^2$

Simplify a bit,

$= \frac{- y + x \frac{d y}{d x}}{y^{2}}$ $= (-y+xdy/dx)/y^2$

Replace $\frac{d y}{d x}$ $dy/dx$ with its equivalent $- \frac{x}{y}$ $-x/y$

$= \frac{- y + x (- \frac{x}{y})}{y^{2}}$ $= (-y+x(-x/y))/y^2$

Simplify again

$= \frac{- y - \frac{x^{2}}{y}}{y^{2}}$ $= (-y-x^2/y)/y^2$

Clear the fraction in the nmumerator

$= \frac{(- y - \frac{x^{2}}{y}) \cdot y}{y^{2} \cdot y}$ $= ((-y-x^2/y)*y)/(y^2*y)$

$= \frac{- y^{2} - x^{2}}{y^{3}}$ $= (-y^2-x^2)/y^3$

Factor out a #-1)

$= \frac{- (y^{2} + x^{2})}{y^{3}}$ $= (-(y^2+x^2))/y^3$

Use the original rfelationship $x^{2} + y^{2} = 90$ $x^2+y^2 = 90$ to finish simplifying.

$= \frac{- 90}{y^{3}}$ $= (-90)/y^3$

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