Find the value of $\frac{d^2y}{dx^2}$ of $x^3 + y^3 = 56$ at $(-2,4)$.
Start by differentiating implicitly both sides of the equation with respect to x. $3x^2 + 3y^2 y' = 0$. Solve for y': $y' = \dfrac{-3x^2}{3y^2} = \dfrac{-x^2}{y^2}$. But this is the first derivative. To find the second derivative, differentiate (implicitly) one more time, now using the quotient rule:
$y'' = \dfrac{(-2x) (y^2) - (-x^2) 2y y'}{(y^2)^2}$. Plug in y':
$y'' = \dfrac{-2xy^2 + 2x^2 y (-x^2/y^2)}{y^4}$.
At this point, we plug in (-2,4). No need to simplify:
$\dfrac{d^2y}{dx^2} |_{(-2,4)} = \dfrac{-2(-2)(4)^2 + 2(-2)^2 (4) (-(-2)^2/4^2)}{4^4}= \dfrac{7}{32}$.
No comments:
Post a Comment