AP Calculus AB / Differentiation Composite, Implicit, and Inverse Functions / Question No. 12

12. The graph of $f (x)$ , shown below, consists of a semicircle and two-line segments. $f^{'} (1) =$

- 1

\frac{- 1}{\sqrt{3}}

\frac{- 3}{\sqrt{8}}

\sqrt{3}

Answer is: option2

\frac{- 1}{\sqrt{3}}

Solution:

Problem: The equation of a circle is $x^{2} + y^{2} = 4$ . Find $\frac{d y}{d x}$ at $x = 1$ .

Step 1: Find $y$ when $x = 1$

$x^{2} + y^{2} = 4$ Substituting $x = 1$ : $1^{2} + y^{2} = 4$ $1 + y^{2} = 4$ $y^{2} = 3$ $y = \sqrt{3}$ Thus, when $x = 1$ , $y = \sqrt{3}$ .

Step 2: Differentiate the equation $x^{2} + y^{2} = 4$ implicitly

Differentiating both sides: $\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (4)$ $2 x + 2 y \frac{d y}{d x} = 0$

Step 3: Solve for $\frac{d y}{d x}$

Solving for $\frac{d y}{d x}$ : $2 y \frac{d y}{d x} = - 2 x$ $\frac{d y}{d x} = - \frac{x}{y}$

Step 4: Substitute $x = 1$ and $y = \sqrt{3}$

Substituting the values into the derivative: $\frac{d y}{d x} = - \frac{1}{\sqrt{3}}$ Thus, the derivative $\frac{d y}{d x}$ at $x = 1$ is $- \frac{1}{\sqrt{3}}$ .

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Step 1: Find $y$ when $x = 1$

Step 2: Differentiate the equation $x^{2} + y^{2} = 4$ implicitly

Step 3: Solve for $\frac{d y}{d x}$

Step 4: Substitute $x = 1$ and $y = \sqrt{3}$

AP Calculus AB Tags

AP Calculus BC Tags

Home Practice Questions AP Calculus AB Differentiation Composite, Implicit, and Inverse Functions Question No. 12

Step 1: Find y when x=1

Step 2: Differentiate the equation x2+y2=4 implicitly

Step 3: Solve for dydx

Step 4: Substitute x=1 and y=3

AP Calculus AB Tags

AP Calculus BC Tags

Step 1: Find $y$ when $x = 1$

Step 2: Differentiate the equation $x^{2} + y^{2} = 4$ implicitly

Step 3: Solve for $\frac{d y}{d x}$

Step 4: Substitute $x = 1$ and $y = \sqrt{3}$