Differentiation: Definition and Fundamental Properties
Unit 2 of AP Calculus AB, worth 10–12% of the exam. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Average versus instantaneous rate of change, the limit definition of the derivative, differentiability and continuity, power, product and quotient rules, derivatives of trig, exponential and log functions.
How this unit is tested
Start by nailing the distinction between average rate of change (a slope of a secant line, computed with (f(b)-f(a))/(b-a)) and instantaneous rate of change (the slope of a tangent line, computed as a limit of average rates as the interval shrinks to a point). Every derivative question on the exam is secretly asking you to recognize which of these two you need.
Next, be fluent with both forms of the limit definition of the derivative: $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ and $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. AP frequently gives you a limit expression and asks you to identify it as f'(a) for some function and value, without ever computing the limit directly — practice recognizing the pattern, not just evaluating it.
Before reaching for a shortcut rule, ask whether the function is even differentiable at the point in question: check that it's continuous there, and that it has no corner, cusp, or vertical tangent. Then move to the mechanical toolkit in this order of complexity: power rule for polynomial pieces, then product and quotient rules whenever two factors are multiplied or divided, then the specific derivative formulas for sine, cosine, tangent, $e^x$, $a^x$, $\ln x$, and $\log_a x$.
Finally, practice combining rules in a single problem, since the exam almost never tests a rule in isolation — a typical free-response or multiple-choice item asks for the derivative of a product of a polynomial and a trig function, or a quotient involving an exponential.
What you have to know
Average rate of change
The average rate of change of f on $[a,b]$ is $\frac{f(b)-f(a)}{b-a}$, the slope of the secant line through $(a,f(a))$ and $(b,f(b))$.
Limit definition of the derivative
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, equivalently $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$; this gives the instantaneous rate of change of f at a point.
Differentiability implies continuity
If f is differentiable at x=a, then f is continuous at x=a. The converse is false: a function can be continuous at a point but fail to be differentiable there (corner, cusp, or vertical tangent).
Power, product, and quotient rules
$\frac{d}{dx}x^n=nx^{n-1}$; $(fg)'=f'g+fg'$; $\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}$.
Derivatives of trigonometric functions
$\frac{d}{dx}\sin x=\cos x$, $\frac{d}{dx}\cos x=-\sin x$, $\frac{d}{dx}\tan x=\sec^2 x$.
Derivatives of exponential and logarithmic functions
$\frac{d}{dx}e^x=e^x$, $\frac{d}{dx}a^x=a^x\ln a$, $\frac{d}{dx}\ln x=\frac{1}{x}$, $\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$.
13 practice questions
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Find the average rate of change of $f(x)=x^3-2x$ on the interval $[1,3]$.
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Answer. 11
Average rate of change is $\frac{f(3)-f(1)}{3-1}$. Compute $f(3)=27-6=21$ and $f(1)=1-2=-1$, so the quotient is $\frac{21-(-1)}{2}=\frac{22}{2}=11$. -
A limit is given as $\lim_{h\to 0}\frac{\sin(\pi/4+h)-\sin(\pi/4)}{h}$. What does this limit represent, and what is its value?
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Answer. It represents f'(π/4) for f(x)=sin x, and equals $\frac{\sqrt{2}}{2}$
This is exactly the limit definition of the derivative applied to f(x)=sin x at a=π/4. Since $\frac{d}{dx}\sin x=\cos x$, the value is $\cos(\pi/4)=\frac{\sqrt{2}}{2}$. -
Use the limit definition to find f'(x) for $f(x)=3x^2+1$.
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Answer. $f'(x)=6x$
Form the difference quotient: $\frac{3(x+h)^2+1-(3x^2+1)}{h}=\frac{6xh+3h^2}{h}=6x+3h$. Taking the limit as h→0 gives $6x$. -
Let $f(x)=|x-2|$. Which statement correctly describes f at x=2?
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Answer. f is continuous at x=2 but not differentiable there
The left and right pieces of |x-2| meet at (2,0) so the function is continuous, but the left-hand slope is -1 and the right-hand slope is +1, giving a corner and no single tangent slope. -
Differentiate $f(x)=5x^4-\frac{3}{x^2}+7$.
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Answer. $f'(x)=20x^3+6x^{-3}$
Rewrite $-\frac{3}{x^2}$ as $-3x^{-2}$. Apply the power rule term by term: $\frac{d}{dx}5x^4=20x^3$, $\frac{d}{dx}(-3x^{-2})=6x^{-3}$, and the derivative of the constant 7 is 0. -
Find the derivative of $f(x)=x^2\sin x$.
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Answer. $f'(x)=2x\sin x+x^2\cos x$
Apply the product rule with $u=x^2$ and $v=\sin x$: $f'=u'v+uv'=2x\sin x+x^2\cos x$. Both terms must be kept and added, not multiplied together. -
Find the derivative of $f(x)=\dfrac{3x+1}{x^2-4}$.
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Answer. $f'(x)=\dfrac{-3x^2-2x-12}{(x^2-4)^2}$
Using the quotient rule with $u=3x+1$, $v=x^2-4$: $f'=\frac{u'v-uv'}{v^2}=\frac{3(x^2-4)-(3x+1)(2x)}{(x^2-4)^2}$. Expanding the numerator: $3x^2-12-(6x^2+2x)=-3x^2-2x-12$. -
Find the derivative of $g(x)=5\sin x-2\cos x$.
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Answer. $g'(x)=5\cos x+2\sin x$
Differentiate term by term using the basic trig derivative rules: $\frac{d}{dx}5\sin x=5\cos x$ and $\frac{d}{dx}(-2\cos x)=-2(-\sin x)=2\sin x$. -
Find the derivative of $f(x)=3e^x-2^x$.
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Answer. $f'(x)=3e^x-2^x\ln 2$
The derivative of $e^x$ is itself, so $\frac{d}{dx}3e^x=3e^x$. For $2^x$, use $\frac{d}{dx}a^x=a^x\ln a$ with a=2, giving $-2^x\ln 2$. -
Find the derivative of $f(x)=\ln x+\log_2 x$.
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Answer. $f'(x)=\dfrac{1}{x}+\dfrac{1}{x\ln 2}$
The natural log derivative is $\frac{1}{x}$. For a log with base 2, use $\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$ with a=2, giving $\frac{1}{x\ln 2}$. -
Find the derivative of $f(x)=x\cos x$ and evaluate f'(0).
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Answer. $f'(x)=\cos x-x\sin x$; f'(0)=1
By the product rule, $f'(x)=(1)\cos x+x(-\sin x)=\cos x-x\sin x$. At x=0, $\cos 0-0\cdot\sin 0=1-0=1$. -
Find the derivative of $f(x)=\dfrac{e^x}{x}$.
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Answer. $f'(x)=\dfrac{e^x(x-1)}{x^2}$
By the quotient rule with $u=e^x$, $v=x$: $f'=\frac{e^x\cdot x-e^x\cdot 1}{x^2}=\frac{e^x(x-1)}{x^2}$. -
Which function fails to be differentiable at x=0 because of a vertical tangent, rather than a corner or a discontinuity?
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Answer. $f(x)=x^{1/3}$
$x^{1/3}$ is continuous everywhere and has no corner, but its derivative $\frac{1}{3}x^{-2/3}$ grows without bound as x→0, producing a vertical tangent line rather than a defined slope.
What people get wrong
- Plugging h=0 directly into the difference quotient before simplifying, which produces 0/0 and stalls the problem. Always algebraically simplify (expand, factor, or rationalize) to cancel the h in the denominator first.
- Assuming continuity guarantees differentiability. Continuity is necessary but not sufficient — functions like $|x|$ or $x^{1/3}$ are continuous but not differentiable at x=0. Always separately check for corners, cusps, or vertical tangents.
- Reversing the order of terms in the quotient rule numerator, giving $fg'-f'g$ instead of $f'g-fg'$. Memorize it as 'low d-high minus high d-low' (using the denominator as 'low') to keep the sign correct.
- Treating the derivative of $a^x$ the same as $e^x$ and forgetting the $\ln a$ factor, or forgetting the extra $\ln a$ in the denominator for $\log_a x$. Only $e^x$ and $\ln x$ are 'clean'; other bases pick up a $\ln a$ factor.
- Confusing average rate of change with instantaneous rate of change on free-response questions — computing a secant slope when the question asks for f'(a), or vice versa. Read whether the question specifies an interval (average) or a single point (instantaneous).
- Applying the product rule by simply multiplying the derivatives of each factor together ($f'g'$) instead of the correct sum $f'g+fg'$. Write out both terms explicitly before simplifying.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.