AP Calculus ABhardmcq1 pt

If $x^2+y^2=16$ defines $y$ implicitly as a function of $x$ , then $\dfrac{d^2y}{dx^2}=$

-\dfrac{x}{y}

-\dfrac{16}{y^3}

\dfrac{16}{y^3}

-\dfrac{1}{y^3}

Explanation

Core Concept

First, $\frac{dy}{dx}=-\frac{x}{y}$ . Differentiate using the quotient rule to get $-\frac{y-x\frac{dy}{dx}}{y^2}$ , then substitute $-\frac{x}{y}$ for $\frac{dy}{dx}$ and use $x^2+y^2=16$ . The result is $-\frac{16}{y^3}$ .

Correct Answer

B $-\dfrac{16}{y^3}$

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If x2+y2=16x^2+y^2=16x2+y2=16 defines yyy implicitly as a function of xxx, then d2ydx2=\dfrac{d^2y}{dx^2}=dx2d2y​=

Explanation

Core Concept

Correct Answer

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