AP Calculus AB Practice Exam — Free

The Fundamental Theorem of Calculus (Part 1) says that if $F(u) = \displaystyle\int_{2}^{u} \sqrt{1 + t^3}\,dt$, then $F'(u) = \sqrt{1 + u^3}$: differentiating an integral with respect to its upper limit returns the integrand evaluated at that limit. Here the upper limit is not $x$ but $u = x^2$, so $g(x) = F(x^2)$ is a composition and the chain rule is required:

Two things happen in that line: the variable $t$ in the integrand is replaced by the upper limit $x^2$ (so $t^3$ becomes $(x^2)^3 = x^6$), and the result is multiplied by the derivative of the upper limit, $2x$. The lower limit 2 is a constant and contributes nothing, since its derivative is zero.

The tempting wrong answer is A, $\sqrt{1 + x^6}$, which substitutes the upper limit correctly but forgets the chain-rule factor $2x$. Option B multiplies by $2x$ but leaves $t^3$ as $x^3$ instead of substituting $x^2$ for $t$, and option D makes both mistakes.

$g'(x) = 2x\sqrt{1 + x^6}$ (option C).

The equation cannot be solved conveniently for $y$, so differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (implicit differentiation). The $xy$ term needs the product rule, and every $y$ term picks up a factor $\dfrac{dy}{dx}$ by the chain rule:

Collect the terms containing $\dfrac{dy}{dx}$ on one side and factor:

Substituting $x = 1$ and $y = 2$ (the point does lie on the curve, since $1+2+4 = 7$):

A negative slope is reasonable: the curve is a tilted ellipse and $(1, 2)$ lies on its upper-right arc, where $y$ decreases as $x$ increases. The most common slip is to write the derivative of $xy$ as $y'$ alone or as $y$ alone instead of $y + xy'$.

$\dfrac{dy}{dx} = -\dfrac{4}{5}$ at $(1, 2)$

(a) The slope at $(1, 2)$ comes from the differential equation: $f'(1) = 2(1)(2)^2 = 8$. The tangent line is $y - 2 = 8(x - 1)$, that is, $y = 8x - 6$. At $x = 1.1$: $f(1.1) \approx 8(1.1) - 6 = 2.8$.

(b) Differentiate $y' = 2xy^2$ with respect to $x$ using the product and chain rules (remember that $y$ depends on $x$):

At $(1, 2)$: $y'' = 2(4) + 8(1)(8) = 72 > 0$. Since $y^2 \ge 0$ and, near $(1,2)$, both $x > 0$ and $y > 0$, the second derivative stays positive, so $f$ is concave up there and its graph lies above its tangent lines. Therefore the tangent-line value 2.8 is an underestimate of $f(1.1)$.

(c) The equation is separable. For $y \ne 0$,

Using $f(1) = 2$: $-\dfrac12 = 1 + C$, so $C = -\dfrac32$. Then

The solution must be defined and differentiable on an interval containing $x = 1$, so the domain is the largest interval around 1 on which the denominator is nonzero: $3-2x^2 > 0$, that is, $-\sqrt{3/2} < x < \sqrt{3/2}$ (about $-1.22 < x < 1.22$). As a comparison, $f(1.1) = \dfrac{2}{3 - 2.42} \approx 3.45$, which is indeed larger than the estimate 2.8, consistent with part (b).

(a) $y = 8x - 6$, $f(1.1) \approx 2.8$; (b) $\dfrac{d^2y}{dx^2} = 2y^2 + 8x^2y^3$, so the estimate is an underestimate; (c) $f(x) = \dfrac{2}{3 - 2x^2}$ on $-\sqrt{3/2} < x < \sqrt{3/2}$.