AP Calculus ABhardmcq1 pt

If $y^2=x$ defines $y$ implicitly as a function of $x$ with $y>0$ , then $\dfrac{d^2y}{dx^2}=$

\dfrac{1}{2y}

-\dfrac{1}{2y^3}

-\dfrac{1}{4y^2}

-\dfrac{1}{4y^3}

Explanation

Core Concept

First, $2y\frac{dy}{dx}=1$ , so $\frac{dy}{dx}=\frac{1}{2y}$ . Differentiate to get $-\frac{1}{2}y^{-2}\frac{dy}{dx}$ , and substitute $\frac{1}{2y}$ to obtain $-\frac{1}{4y^3}$ .

Correct Answer

D $-\dfrac{1}{4y^3}$

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Explanation

Core Concept

Correct Answer

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