Unit 3 practice test: Differentiation: Composite, Implicit, and Inverse Functions

Treat $y$ as a function of $x$. Then $\frac{d}{dx}\left(x^2 + y^2\right)$$= 2x + 2y\frac{dy}{dx}$$= 0$, so $\frac{dy}{dx} = -\frac{x}{y}$.

• The derivative of $25$ should be $25$: 25 is a constant, so its derivative is 0; that part was right.
• The derivative of $x^2$ should be $2x\frac{dy}{dx}$: Differentiating with respect to $x$, the derivative of $x^2$ is just $2x$.
• Both sides should be divided by $2$ before differentiating: Dividing by 2 changes nothing important; the missing $\frac{dy}{dx}$ is the error.
• Correct: $y$ depends on $x$, so $\frac{d}{dx}y^2 = 2y\frac{dy}{dx}$: The chain rule applies to every term that holds $y$; the student dropped the factor $\frac{dy}{dx}$.

With $\frac{dy}{dx} = -\frac{x}{y}$, a zero denominator and a nonzero numerator mean no slope: the tangent line at $(-5, 0)$ is vertical, as the graph shows.

• Correct: It is undefined, because $y = 0$ makes the denominator 0: $-\frac{x}{y} = -\frac{-5}{0}$ has no value; the tangent line there is vertical.
• It equals 0, because $y = 0$: A zero denominator does not make a fraction 0. The slope is 0 where the numerator $x$ is 0, at $(0, \pm 5)$.
• It equals $-5$, the value of $x$: The slope is $-\frac{x}{y}$, not $x$.
• It equals 1, because the circle is round: The slope changes from point to point; at $(-5, 0)$ the formula gives a zero denominator.

Solve $-\frac{x}{y} = \frac{3}{4}$ on the circle: $(3, -4)$ works, since $-\frac{3}{-4} = \frac{3}{4}$ and $9 + 16 = 25$.

• $(3, 4)$: $-\frac{3}{4}$: the sign is wrong.
• Correct: $(3, -4)$: $-\frac{3}{-4} = \frac{3}{4}$.
• $(4, 3)$: $-\frac{4}{3}$ at this point.
• $(-4, -3)$: $-\frac{-4}{-3} = -\frac{4}{3}$ at this point.

$k'(x)$$= \frac{g'(x)}{1 + [g(x)]^2}$, so $k'(2)$$= \frac{-4}{1 + 0}$$= -4$.

• $1$: This is $\frac{1}{1 + [g(2)]^2}$ without the inside derivative $g'(2) = -4$.
• Correct: $-4$: $\frac{g'(2)}{1 + [g(2)]^2}$$= \frac{-4}{1}$.
• $-\frac{4}{17}$: This puts $g'(2)$ in the denominator. The denominator uses the value $g(2) = 0$.
• $4$: The sign of $g'(2) = -4$ carries through.

$m'(x)$$= -\frac{g'(x)}{\sqrt{1 - [g(x)]^2}}$, so $m'(4)$$= -\frac{6}{\sqrt{1 - \frac{1}{4}}}$$= -\frac{6}{\sqrt{3}/2}$$= -4\sqrt{3}$.

• $4\sqrt{3}$: The derivative of $\arccos u$ carries a minus sign.
• $-\frac{2\sqrt{3}}{3}$: This is $-\frac{1}{\sqrt{3}/2}$ without the inside derivative $g'(4) = 6$.
• Correct: $-4\sqrt{3}$: $-\frac{6}{\sqrt{3}/2}$$= -\frac{12}{\sqrt{3}}$$= -4\sqrt{3}$.
• $-\frac{24}{5}$: This uses $1 + u^2$ in the denominator, the arctangent pattern, instead of $\sqrt{1 - u^2}$.

The derivative of $\arcsin u$ is $\frac{1}{\sqrt{1 - u^2}}$, defined for $-1 < u < 1$. At $x = 3$, $u = g(3) = 1$ and the denominator is 0.

• $g'(3) = 2$ is greater than 1, outside the domain of arcsine: The domain condition is on the inside value $g(3)$, not on $g'(3)$.
• $\arcsin 1$ is not defined: $\arcsin 1 = \frac{\pi}{2}$ is defined; it is the derivative that fails at $u = 1$.
• The table gives no value of $g$ near 3: The formula itself breaks: its denominator $\sqrt{1 - [g(3)]^2}$ is 0.
• Correct: $g(3) = 1$, so $\sqrt{1 - [g(3)]^2} = 0$ and the formula divides by 0: The derivative of $\arcsin u$ exists only for $-1 < u < 1$; here $u = 1$.