MATH 201 — Exam II Review

Exam II Review — The Derivative, Start to Finish
MATH 201  Calculus I  OpenStax Calculus Volume 1, §3.1–3.9

How to use this note. This is a consolidation, not new material. Everything here has appeared since Oct 5. The exam asks you to do four things: compute a derivative from the definition, apply the rules to a function you have never seen, use a derivative to answer a question about something (a rate, a tangent line, a direction), and know which rule the problem is secretly asking for. Each section below is organized that way — what you must know cold, the move that uses it, and the error that costs the most points.

What is covered.

§	Topic	What you must be able to do
3.1–3.2	The derivative	Compute $f'$f′ from the limit definition; say where $f$f fails to be differentiable
3.3	Rules	Power, product, quotient
3.4	Rates of change	Position $\to$→ velocity $\to$→ acceleration; interpret the sign
3.5	Trigonometric	The six derivatives, cold
3.6	Chain rule	Including a chain inside a chain
3.7	Inverse functions	$(f^{-1})'(b) = 1/f'(a)$(f−1)′(b)=1/f′(a); inverse trig derivatives
3.8	Implicit differentiation	Differentiate $y$y terms with $dy/dx$dy/dx; solve; tangent lines
3.9	Exponential, logarithmic	$e^{x}$ex, $a^{x}$ax, $\ln x$lnx, $\log_{a}x$loga​x; logarithmic differentiation

Related rates (§4.1) is not on this exam.

The derivative from the definition (§3.1–3.2)

Know cold. The two equivalent forms

$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}, \qquad f'(a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}.$f′(x)=h→0lim​hf(x+h)−f(x)​,f′(a)=x→alim​x−af(x)−f(a)​.

And the one-way implication: differentiable $\Rightarrow$⇒ continuous, never the reverse.

The move. With a square root, multiply by the conjugate; with a fraction, combine over a common denominator. In both cases the goal is identical: produce a factor of $h$h in the numerator that cancels the $h$h underneath. Until that $h$h cancels, the limit is $0/0$0/0 and you have not finished.

Trap. Cancelling $h$h before rationalising. If you write $\frac{\sqrt{x+h}-\sqrt{x}}{h}$hx+h​−x​​ and simply “cancel,” you have invented algebra. Rationalise first.

Trap. A function can be continuous at a point and still fail to be differentiable there. At a junction of a piecewise function you must check both one-sided limits of the difference quotient — not just that the two pieces meet.

Example 1 (The definition, with a fraction). For $f(x) = \dfrac{1}{x+3}$f(x)=x+31​,

$\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left(\frac{1}{x+h+3}-\frac{1}{x+3}\right) = \frac{1}{h}\cdot\frac{(x+3)-(x+h+3)}{(x+h+3)(x+3)} = \frac{-h}{h\,(x+h+3)(x+3)} ,$hf(x+h)−f(x)​=h1​(x+h+31​−x+31​)=h1​⋅(x+h+3)(x+3)(x+3)−(x+h+3)​=h(x+h+3)(x+3)−h​,

and now the $h$h cancels, leaving $f'(x) = -\dfrac{1}{(x+3)^{2}}$f′(x)=−(x+3)21​.

The rules (§3.3, §3.6)

Know cold.

$(fg)' = f'g + fg', \qquad \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^{2}}, \qquad \bigl(f(g(x))\bigr)' = f'\bigl(g(x)\bigr)\cdot g'(x).$(fg)′=f′g+fg′,(gf​)′=g2f′g−fg′​,(f(g(x)))′=f′(g(x))⋅g′(x).

The move. Before differentiating, say out loud what the function is: a product? a quotient? a composition? Most lost points on this exam will come from applying a correct rule to a function of the wrong shape.

Trap. The quotient rule is not symmetric. $f'g - fg'$f′g−fg′, in that order. Swapping them negates your answer, and the rest of the problem then looks right.

Trap. The chain rule’s inner derivative is the single most-forgotten factor in the course. $\frac{d}{dx}\sin(5x) = 5\cos(5x)$dxd​sin(5x)=5cos(5x), not $\cos(5x)$cos(5x).

Example 2 (A chain inside a chain). $y = \left(\tan(3x)\right)^{4}$y=(tan(3x))4. Outermost is the fourth power, then $\tan$tan, then $3x$3x:

$y' = 4\left(\tan 3x\right)^{3}\cdot \sec^{2}(3x)\cdot 3 = 12\left(\tan 3x\right)^{3}\sec^{2}(3x).$y′=4(tan3x)3⋅sec2(3x)⋅3=12(tan3x)3sec2(3x).

Three layers, three factors. Count your layers before you start.

Trigonometric derivatives and rates of change (§3.5, §3.4)

Know cold. All six, including signs:

$(\sin)' = \cos, \quad (\cos)' = -\sin, \quad (\tan)' = \sec^{2},$(sin)′=cos,(cos)′=−sin,(tan)′=sec2, $(\cot)' = -\csc^{2}, \quad (\sec)' = \sec\tan, \quad (\csc)' = -\csc\cot.$(cot)′=−csc2,(sec)′=sectan,(csc)′=−csccot.

Every “co” function carries a minus sign. That is the whole pattern.

The move. For motion on a line: $v(t) = s'(t)$v(t)=s′(t) and $a(t) = v'(t) = s''(t)$a(t)=v′(t)=s′′(t). To find when the object is at rest, solve $v(t)=0$v(t)=0; to find which way it is moving, take the sign of $v$v on each interval between those times.

Trap. “Moving in the negative direction” is $v < 0$v<0, not $a < 0$a<0. Acceleration negative with velocity positive means slowing down while still moving forward.

Trap. Speed is $\left\lvert v \right\rvert$∣v∣, not $v$v. A question about speed increasing is asking whether $v$v and $a$a have the same sign.

Example 3 (Reading the signs). If $s(t) = t^{3}-6t^{2}+9t$s(t)=t3−6t2+9t, then $v(t) = 3t^{2}-12t+9 = 3(t-1)(t-3)$v(t)=3t2−12t+9=3(t−1)(t−3) and $a(t) = 6t-12$a(t)=6t−12. At rest at $t=1,3$t=1,3. On $(1,3)$(1,3), $v<0$v<0 — moving backwards. At $t=2$t=2, $v<0$v<0 and $a=0$a=0; just after, $a>0$a>0 while $v<0$v<0, so the particle is slowing down.

Inverses and implicit differentiation (§3.7, §3.8)

Know cold. If $f(a)=b$f(a)=b and $f'(a) \neq 0$f′(a)=0, then

$\left(f^{-1}\right)'(b) = \frac{1}{f'(a)} = \frac{1}{f'\!\left(f^{-1}(b)\right)} .$(f−1)′(b)=f′(a)1​=f′(f−1(b))1​.

And the inverse trig derivatives:

$\left(\arcsin x\right)' = \frac{1}{\sqrt{1-x^{2}}}, \qquad \left(\arccos x\right)' = -\frac{1}{\sqrt{1-x^{2}}}, \qquad \left(\arctan x\right)' = \frac{1}{1+x^{2}} .$(arcsinx)′=1−x2​1​,(arccosx)′=−1−x2​1​,(arctanx)′=1+x21​.

The move (inverses). Three steps, always. (1) Check $f$f is one-to-one, usually by showing $f' > 0$f′>0 everywhere or $f' < 0$f′<0 everywhere. (2) Find the point $a$a with $f(a)=b$f(a)=b — by inspection, not by finding a formula for $f^{-1}$f−1. (3) Answer $1/f'(a)$1/f′(a).

The move (implicit). Differentiate both sides with respect to $x$x, attaching $\frac{dy}{dx}$dxdy​ every time a $y$y is differentiated; collect the $\frac{dy}{dx}$dxdy​ terms on one side; factor and divide.

Trap. The single most expensive error in §3.7: evaluating $f'$f′ at $b$b instead of at $a$a. $\left(f^{-1}\right)'(b) = 1/f'(a)$(f−1)′(b)=1/f′(a), and $a$a is the input whose output is $b$b.

Trap. In §3.8, a term like $xy$xy needs the product rule and then a $\frac{dy}{dx}$dxdy​: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$dxd​(xy)=y+xdxdy​. Writing just $\frac{dy}{dx}$dxdy​ loses the problem.

Example 4 (The point, not the formula). Let $f(x) = x^{3}+4x+2$f(x)=x3+4x+2. Then $f'(x) = 3x^{2}+4 > 0$f′(x)=3x2+4>0, so $f$f is one-to-one. To find $\left(f^{-1}\right)'(7)$(f−1)′(7): solve $a^{3}+4a+2 = 7$a3+4a+2=7, i.e. $a^{3}+4a-5=0$a3+4a−5=0, and $a=1$a=1 works. Since $f'(1) = 7$f′(1)=7, the answer is $\frac{1}{7}$71​. No formula for $f^{-1}$f−1 was ever needed.

Exponentials and logarithms (§3.9)

Know cold.

$\left(e^{x}\right)' = e^{x}, \quad \left(a^{x}\right)' = a^{x}\ln a, \quad \left(\ln x\right)' = \frac{1}{x}, \quad \left(\log_{a}x\right)' = \frac{1}{x\ln a}.$(ex)′=ex,(ax)′=axlna,(lnx)′=x1​,(loga​x)′=xlna1​.

The move. Ask what is varying. If only the exponent varies, it is $a^{x}$ax. If only the base varies, it is the power rule. If both vary — $\left(\text{stuff}\right)^{\text{stuff}}$(stuff)stuff — neither rule applies and you must use logarithmic differentiation: take $\ln$ln of both sides, differentiate implicitly, then multiply back by $y$y.

Trap. $\frac{d}{dx}x^{x}$dxd​xx is neither $x\cdot x^{x-1}$x⋅xx−1 nor $x^{x}\ln x$xxlnx. Both rules are wrong here, and each is wrong in a way that looks plausible.

Example 5 (Logarithmic differentiation). $y = \left(\ln x\right)^{x}$y=(lnx)x for $x>1$x>1. Then $\ln y = x\ln(\ln x)$lny=xln(lnx), so

$\frac{y'}{y} = \ln(\ln x) + x\cdot\frac{1}{\ln x}\cdot\frac{1}{x} = \ln(\ln x) + \frac{1}{\ln x},$yy′​=ln(lnx)+x⋅lnx1​⋅x1​=ln(lnx)+lnx1​,

and $y' = \left(\ln x\right)^{x}\left(\ln(\ln x) + \frac{1}{\ln x}\right)$y′=(lnx)x(ln(lnx)+lnx1​).

Practice problems

Deliberately not shaped like the exam — the point is to stretch the same skills over different ground. Work them without notes first.

A. Say it exactly

1. Write the limit definition of $f'(a)$f′(a) in both forms. State the relationship between differentiability and continuity, and give a function that shows the converse fails.

2. Name the three ways a function can fail to be differentiable at a point, and sketch one example of each.

B. Find the error

Each is wrong. Say exactly where.

1. “$\frac{d}{dx}\left[x^{2}\sin x\right] = 2x\cos x$dxd​[x2sinx]=2xcosx.”

2. “$\frac{d}{dx}\cos(x^{3}) = -\sin(3x^{2})$dxd​cos(x3)=−sin(3x2).”

3. “Since $f(x)=\left\lvert x \right\rvert$f(x)=∣x∣ is continuous at $0$0, it is differentiable at $0$0.”

4. “If $f(x) = x^{5}+x$f(x)=x5+x then $\left(f^{-1}\right)'(2) = \frac{1}{f'(2)} = \frac{1}{81}$(f−1)′(2)=f′(2)1​=811​.”

C. From the definition

1. Use the limit definition to find $f'(x)$f′(x) for $f(x) = \sqrt{3x-2}$f(x)=3x−2​.

2. Use the limit definition to find $f'(x)$f′(x) for $f(x) = \dfrac{2}{x-1}$f(x)=x−12​.

3. Let $g(x) = x^{2}$g(x)=x2 for $x \le 1$x≤1 and $g(x) = 2x-1$g(x)=2x−1 for $x>1$x>1. Is $g$g continuous at $1$1? Differentiable at $1$1? Justify both with one-sided limits.

D. Differentiate

1. $y = \dfrac{x^{2}-1}{x^{2}+1}$y=x2+1x2−1​  (b) $y = x^{3}\sec x$y=x3secx  (c) $y = \sqrt3$y=3​

2. $y = \sin^{2}\left(e^{4x}\right)$y=sin2(e4x)  (b) $y = \dfrac{\arctan x}{x}$y=xarctanx​  (c) $y = 5^{x^{2}}$y=5x2

3. $y = x^{\ln x}$y=xlnx,  $x>0$x>0. (Which method is forced here, and why?)

E. Rates of change

1. A particle has position $s(t) = 2t^{3}-15t^{2}+36t$s(t)=2t3−15t2+36t for $t \ge 0$t≥0. Find $v$v and $a$a; find all times at rest; find the intervals where it moves backwards; and decide whether it is speeding up or slowing down at $t=3$t=3.

2. A ball’s height is $h(t) = 48t - 16t^{2}$h(t)=48t−16t2 feet. What is its velocity at the instant it returns to the ground, and how does that compare with its initial velocity?

F. Inverses and implicit

1. Let $f(x) = x^{3}+2x+1$f(x)=x3+2x+1. Explain why $f^{-1}$f−1 exists, then compute $\left(f^{-1}\right)'(4)$(f−1)′(4).

2. Find $\dfrac{dy}{dx}$dxdy​ for $x^{3}+y^{3} = 6xy$x3+y3=6xy, and the tangent line at $(3,3)$(3,3).

3. Find $\dfrac{dy}{dx}$dxdy​ for $\sin(xy) = x + y$sin(xy)=x+y.

4. Differentiate $y = \arcsin(3x)$y=arcsin(3x) and $y = x\arctan\left(x^{2}\right)$y=xarctan(x2).

G. Mixed — name the rule first

1. For each, say which rule or rules the problem requires before differentiating: (a) $e^{x}\ln x$exlnx;  (b) $\left(x^{2}+1\right)^{10}$(x2+1)10;  (c) $\dfrac{\ln x}{x^{2}}$x2lnx​;  (d) $\left(\sin x\right)^{x}$(sinx)x;  (e) $\arctan\left(e^{x}\right)$arctan(ex).

2. Find all points on the curve $y = x^{3}-3x$y=x3−3x where the tangent line is horizontal.

3. Find the equation of the tangent line to $y = xe^{x}$y=xex at $x=0$x=0.