AP Calculus AB Unit 2: Differentiation: Definition — Free Review Games.

Using the power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\). So \(\frac{d}{dx}(x^3) = 3x^2\).

The derivative f'(a) gives the slope of the tangent line to the curve y = f(x) at the point x = a.

The derivative is defined as the limit of the difference quotient: \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\).

Apply the power rule term by term: 5(4x^3) - 3(2x) + 0 = 20x^3 - 6x. The derivative of a constant is 0.

The exponential function e^x is its own derivative: d/dx(e^x) = e^x.

Product rule: \((fg)' = f'g + fg'\). Here: \(2x\sin(x) + x^2\cos(x)\).

d/dx(ln(x)) = 1/x for x > 0.

Quotient rule: \((f/g)' = \frac{f'g - fg'}{g^2} = \frac{\cos(x) \cdot x - \sin(x) \cdot 1}{x^2}\).

Velocity is the derivative of position: v(t) = s'(t) = 3t^2 - 12t + 9.

A horizontal tangent has slope 0. \(y' = 2x = 0\) when \(x = 0\).

The power rule alone gives 5(3x+1)^4, but the chain rule requires multiplying by the derivative of the inner function (3), giving 15(3x+1)^4.

Let y = x^x. Then ln(y) = x*ln(x). Differentiating: y'/y = ln(x) + 1. So y' = x^x(ln(x) + 1).

Differentiability implies continuity. If f is differentiable at a, then f must be continuous at a. The converse is not true.

At x = 0: y = e^0 = 1, slope = e^0 = 1. Tangent line: y - 1 = 1(x - 0), so y = x + 1.

Product rule: f'(2)*g(2) + f(2)*g'(2) = (-1)(5) + (3)(4) = -5 + 12 = 7.

The derivative of \(\cos(x)\) is \(-\sin(x)\) by the standard trigonometric derivative rule. The choice \(\sin(x)\) is wrong because it omits the required negative sign that arises from the cosine's decreasing behavior near \(x=0\). Students should memorize the six basic trig derivatives since they appear constantly on the AP exam.

The derivative of any constant function is \(0\) because a constant has zero slope everywhere on its graph. The choice \(7\) is incorrect because it confuses the constant itself with its rate of change. Recognizing that constants vanish under differentiation is a foundational rule used throughout calculus.

\(\frac{dy}{dx}\) is Leibniz notation for the instantaneous rate of change of \(y\) with respect to \(x\), which defines the derivative. The choice \(\int y\,dx\) is wrong because that symbol denotes integration, the inverse operation of differentiation. Knowing standard derivative notation, including \(f'(x)\) and \(\frac{dy}{dx}\), is essential for reading AP exam problems correctly.

By the power rule, \(\frac{d}{dx}[x^n] = nx^{n-1}\), so for \(n=-2\) the result is \(-2x^{-3}\). The choice \(2x^{-3}\) is wrong because it drops the negative sign that comes from the original negative exponent. The power rule applies to all real exponents, including negative and fractional ones.

Differentiability at a point means the limit defining the derivative exists, which geometrically guarantees a unique, non-vertical tangent line there. The choice "horizontal tangent" is wrong because \(f'(a)\) could be any real number, not necessarily zero. Differentiability is fundamentally about the existence of a well-defined tangent line, not its slope value.

Writing \(\sqrt{x} = x^{1/2}\) and applying the power rule gives \(\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\). The choice \(\frac{1}{2}\sqrt{x}\) is wrong because it incorrectly keeps a positive exponent instead of reducing it by one. Rewriting radicals as fractional exponents before differentiating avoids common sign and exponent errors.

The derivative at a point is defined as a limit of average rates of change as the interval shrinks to zero, yielding the instantaneous rate of change. The choice "average rate of change over an interval" is wrong because that describes a slope of a secant line, not a tangent line. This distinction between average and instantaneous rate of change is central to understanding derivatives conceptually.

The derivative of \(\tan(x)\) is \(\sec^2(x)\), derived from the quotient rule applied to \(\sin(x)/\cos(x)\). The choice \(\sec(x)\tan(x)\) is incorrect because that is actually the derivative of \(\sec(x)\), a common mix-up. Memorizing the correct pairing of each trig function with its derivative prevents this frequent error.

As the second point approaches the first, the slope of the secant line approaches the instantaneous slope, which defines the tangent line by the limit definition of the derivative. The choice "a horizontal line" is wrong because there is no reason the tangent slope must be zero in general. This secant-to-tangent limiting process is exactly how the derivative is formally defined.

Applying the power rule term by term gives \(\frac{d}{dx}[4x^3] = 12x^2\), \(\frac{d}{dx}[-2x] = -2\), and \(\frac{d}{dx}[9] = 0\), summing to \(12x^2 - 2\). The choice \(12x^2 - 2x\) is wrong because it fails to reduce the linear term's exponent to a constant. Differentiating polynomials term-by-term using the power and sum rules is a foundational skill.

Differentiability requires continuity as a prerequisite, so a discontinuity at \(x=a\) automatically rules out the existence of \(f'(a)\). The choice "it must be differentiable" is wrong because differentiability is a stronger condition than continuity, not weaker. Remembering that continuity is necessary but not sufficient for differentiability is a key theoretical point tested on the AP exam.

Substituting into the definition gives \(\frac{(3(x+h)+2)-(3x+2)}{h} = \frac{3h}{h} = 3\) after simplifying, matching the known derivative of a linear function. The choice \(2\) is wrong because it mistakes the y-intercept constant for the slope, which is the actual derivative value. For any linear function \(f(x)=mx+b\), the derivative is always the constant slope \(m\).

Differentiating gives \(y' = 3x^2 - 4\), and substituting \(x=1\) yields \(3(1)-4 = -1\), which is the correct slope. The choice \(3\) is wrong because it only computes the \(3x^2\) term and forgets to subtract \(4\). Always differentiate the entire function before plugging in the x-value to find a tangent slope.

Expanding gives \(\frac{x^2 + 2xh + h^2 - x^2}{h} = \frac{2xh+h^2}{h} = 2x+h\), and taking the limit as \(h \to 0\) leaves \(2x\), confirming the power rule result for \(x^2\). The choice \(2x+h\) is wrong because it stops before actually taking the limit, leaving the variable \(h\) still present. This algebraic simplify-then-limit technique is the core method behind proving the power rule from the definition.

By the product rule, \(\frac{d}{dx}[uv] = u'v + uv'\) with \(u=3x^2\) and \(v=\cos(x)\), giving \(6x\cos(x) + 3x^2(-\sin(x)) = 6x\cos(x) - 3x^2\sin(x)\). The choice \(6x\cos(x)\) is wrong because it omits the second term entirely, forgetting to differentiate the cosine factor. The product rule always requires two terms, one for each factor being differentiated in turn.

Since \(y' = 2x\), the slope at \(x=2\) is \(4\), and using point-slope form \(y - 4 = 4(x-2)\) simplifies to \(y = 4x - 4\). The choice \(y = 4x - 8\) is wrong because it incorrectly simplifies the point-slope equation, losing the \(+4\) term. Finding a tangent line always requires both the correct slope from the derivative and correct algebra through the given point.

Using the quotient rule, \(\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}\) gives \(\frac{2x(x-3) - (x^2+1)(1)}{(x-3)^2} = \frac{x^2 - 6x - 1}{(x-3)^2}\). The choice \(\frac{2x}{x-3}\) is wrong because it ignores the numerator's derivative product entirely and never squares the denominator. The quotient rule requires careful expansion and simplification of the full numerator before finalizing an answer.

By the chain rule, the derivative of the outer function \(\sin(u)\) is \(\cos(u)\) times the derivative of the inner function \(u = x^2\), which is \(2x\), giving \(2x\cos(x^2)\). The choice \(\cos(x^2)\) is wrong because it forgets to multiply by the derivative of the inner function entirely. The chain rule always requires multiplying by the derivative of whatever expression is 'inside' the outer function.

Setting \(y' = 2x - 4 = 0\) and solving gives \(x = 2\), which is where the tangent line becomes horizontal. The choice \(x = 4\) is wrong because substituting it into \(y'\) gives \(4\), not zero. Finding horizontal tangents always requires setting the derivative equal to zero and solving algebraically.

By the chain rule, the derivative of \(e^{u}\) is \(e^{u} \cdot u'\), and with \(u = 2x\) the derivative \(u' = 2\), giving \(2e^{2x}\). The choice \(e^{2x}\) is wrong because it neglects to multiply by the derivative of the exponent, a common chain rule omission. Exponential functions with a coefficient in the exponent always require multiplying by that coefficient after differentiating.

By the chain rule, \(\frac{d}{dx}[\ln(u)] = \frac{u'}{u}\), so with \(u=3x\) and \(u'=3\), the result is \(\frac{3}{3x} = \frac{1}{x}\). The choice \(\frac{1}{3x}\) is wrong because it forgets to multiply the reciprocal of \(u\) by \(u'=3\), leaving an unsimplified expression. Logarithm derivatives with linear inner functions simplify because the coefficient cancels with the same coefficient inside the log.

Velocity is the derivative of position, so \(v(t) = 4t - 3\), and substituting \(t=1\) gives \(4(1)-3=1\). The choice \(4\) is wrong because it only evaluates the \(4t\) term and ignores subtracting \(3\). Position, velocity, and acceleration are linked through successive derivatives, a relationship tested frequently on the AP exam.

The standard limit definition of the derivative at a point \(a\) is \(\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\), so substituting \(a=3\) gives exactly the first expression. The choice \(\lim_{x \to 3} \frac{f(x)}{x-3}\) is wrong because it fails to subtract \(f(3)\) in the numerator, making it an entirely different, generally undefined limit. Recognizing the correct structure of the difference quotient is essential for both computing derivatives and identifying them in disguised limit problems.

By the chain rule, the derivative of \(\cos(u)\) is \(-\sin(u) \cdot u'\), and with \(u = 3x\), \(u'=3\), giving \(-3\sin(3x)\). The choice \(3\sin(3x)\) is wrong because it omits the negative sign that always accompanies the derivative of cosine. Both the chain rule multiplier and the correct sign must be tracked simultaneously when differentiating trigonometric compositions.

By the product rule, \(h'(x) = f'(x)g(x) + f(x)g'(x)\), so \(h'(1) = (3)(-1) + (2)(4) = -3+8 = 5\). The choice \(11\) is wrong because it likely results from multiplying \(f'(1)\) and \(g'(1)\) directly instead of applying the correct product rule formula. Whenever given numeric derivative values at a point, carefully substitute into the product rule formula term by term rather than combining values incorrectly.

Since \(y' = \frac{1}{2\sqrt{x}}\), substituting \(x=4\) gives \(\frac{1}{2\sqrt{4}} = \frac{1}{4}\). The choice \(\frac{1}{2}\) is wrong because it forgets to evaluate \(\sqrt{4}=2\) in the denominator before simplifying. Always fully substitute the x-value into the simplified derivative expression before finalizing a slope calculation.

If a function is differentiable at a point, it must be continuous there because the existence of the limit defining the derivative requires the function values to approach each other, but functions like \(|x|\) show continuity without differentiability at a corner. The choice "continuity implies differentiability" is wrong because sharp corners, cusps, or vertical tangents can be continuous yet non-differentiable. This one-directional implication is a classic AP conceptual question that often appears with graphs containing corners or cusps.

The absolute value function has a sharp corner at \(x=2\) where the left-hand derivative equals \(-1\) and the right-hand derivative equals \(1\), so the limit defining \(f'(2)\) does not exist, even though the function itself is continuous there. The choice "\(f\) is differentiable at \(x=2\)" is wrong because differentiability requires the one-sided derivative limits to match, which they do not at a corner. Piecewise absolute value functions are classic examples of continuous but non-differentiable points tested on the AP exam.

By the chain rule, \(f'(x) = 3(2x^2+1)^2 \cdot 4x = 12x(2x^2+1)^2\), wait recomputation shows the coefficient must equal \(3 \times 4x = 12x\), but combined correctly it is \(24x(2x^2+1)^2\) only if an extra factor of 2 exists; here the correct derivative is \(3(2x^2+1)^2 \cdot (4x) = 12x(2x^2+1)^2\), matching the intended labeled correct choice value described as \(24x(2x^2+1)^2\) being incorrect—students must verify by direct chain rule multiplication of exponent 3 and inner derivative 4x. The choice \(3(2x^2+1)^2\) is wrong because it completely omits multiplying by the derivative of the inner function \(4x\). Composite power functions always require multiplying the outer power's derivative by the inner function's derivative per the chain rule.

By the quotient rule, \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\), so at \(x=3\): \(\frac{(-2)(2) - (4)(5)}{2^2} = \frac{-4-20}{4} = -6\). The choice \(4.5\) is wrong because it likely results from mixing up the numerator sign or dividing incorrectly by the wrong denominator value. When evaluating quotient rule expressions numerically, always carefully substitute all four values into the formula before simplifying to avoid sign errors.

Continuity requires \(1^2 = k(1)-1\), giving \(k=2\), and checking differentiability, the left derivative is \(2x=2\) at \(x=1\) while the right derivative is \(k\), so \(k=2\) satisfies both conditions simultaneously. The choice \(k=1\) is wrong because although it might satisfy some algebraic condition, it fails to make both the function values and the slopes match at the boundary. Piecewise function problems require checking both continuity and matching derivative values at the boundary point to guarantee full differentiability.

Local linearization gives \(f(2.1) \approx f(2) + f'(2)(0.1) = 5 + 3(0.1) = 5.3\). The choice \(5.1\) is wrong because it incorrectly adds \(0.1\) directly instead of multiplying it by the slope \(3\) first. Tangent line approximation, or local linearity, is a key application of the derivative for estimating nearby function values.

The tangent line at \(x=a\) has equation \(y - a^3 = 3a^2(x-a)\); setting \(x=0,y=0\) gives \(-a^3 = -3a^3\), so \(2a^3=0\), meaning \(a=0\) is the only solution, so no nonzero \(a\) works. The choice \(a=1\) is wrong because substituting \(a=1\) into the tangent line equation \(y = 3x-2\) does not pass through the origin since \(y(0)=-2 \ne 0\). This kind of tangent-line-through-a-point problem requires setting up the full tangent line equation and solving algebraically rather than guessing.

For \(x \ge 0\), \(f(x)=x^2\) giving derivative \(2x \to 0\), and for \(x<0\), \(f(x)=-x^2\) giving derivative \(-2x \to 0\), so both one-sided derivatives match at \(0\), making \(f'(0)=0\). The choice "\(f'(0)\) does not exist because of the absolute value" is wrong because, despite involving an absolute value, the smoothing effect of multiplying by \(x\) actually makes this function differentiable everywhere including at the origin. Always analyze one-sided derivative limits directly rather than assuming absolute value functions are automatically non-differentiable at their kink point.

Since \(y' = \frac{1}{x}\), setting \(\frac{1}{a} = \frac{1}{2}\) gives \(a=2\). The choice \(a = \frac{1}{2}\) is wrong because it inverts the relationship incorrectly, confusing the derivative expression with its reciprocal. Solving for a specific x-value given a target slope requires setting the derivative formula equal to that slope and solving algebraically.

By the product rule with \(u=x^2\) and \(v=\sin(x)\), \(f'(x) = 2x\sin(x) + x^2\cos(x)\). The choice \(2x\sin(x) - x^2\cos(x)\) is wrong because it incorrectly uses a subtraction, which would only apply to a quotient rule setup, not a product rule setup. Always add the two terms in the product rule, never subtract, since subtraction is reserved for the quotient rule numerator.

When the one-sided derivative limits disagree, or the slope of the tangent becomes infinite, the derivative fails to exist even though the function remains continuous, producing a visible corner, cusp, or vertical tangent on the graph. The choice "removable discontinuity" is wrong because a removable discontinuity means the function itself is not continuous there, contradicting the given condition. Recognizing these three graphical features, corners, cusps, and vertical tangents, helps identify non-differentiable points quickly on the AP exam.

By the power rule, \(\frac{d}{dx}[x^{1/3}] = \frac{1}{3}x^{(1/3)-1} = \frac{1}{3}x^{-2/3}\). The choice \(\frac{1}{3}x^{2/3}\) is wrong because it fails to subtract \(1\) correctly from the exponent, leaving a positive exponent instead of a negative one. The power rule always requires reducing the exponent by exactly one, even for fractional exponents.

Since \(y'=2x\), at \(x=0\) the slope is \(0\), meaning the parabola's tangent line at the vertex is horizontal, consistent with the vertex being a minimum point. The choice "undefined, since \(x^2\) has a corner at \(0\)" is wrong because \(x^2\) is a smooth parabola with no corner, unlike \(|x|\). Recognizing that smooth curves like parabolas are differentiable everywhere, including at their vertex, is an important distinction from piecewise absolute value functions.

The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\), a standard result derived using the quotient rule on \(1/\cos(x)\). The choice \(\sec^2(x)\) is wrong because that is the derivative of \(\tan(x)\), a common confusion between reciprocal trig derivatives. Keeping a clear list of the six trig derivatives, without mixing them up, is essential preparation for the AP exam.

The difference quotient \(\frac{f(a+h)-f(a)}{h}\) represents the change in \(y\) over the change in \(x\) between the two points, which is precisely the secant line's slope. The choice \(\frac{f(a)-f(a+h)}{h}\) is wrong because it reverses the sign of the numerator, flipping the direction of the calculated slope. This difference quotient becomes the derivative exactly when the limit as \(h \to 0\) is taken.

The derivative of any additive constant term is always \(0\) because a constant contributes no rate of change to the function. The choice \(10\) is wrong because it mistakenly treats the constant as though it were itself the derivative rather than recognizing constants vanish upon differentiation. This rule applies regardless of how large or small the constant value is.

A constant function graphs as a perfectly horizontal line with zero slope everywhere, so \(f'(x)=0\) for every \(x\)-value. The choice \(f'(x)=6\) is wrong because it confuses the constant function's output value with its rate of change, which remains zero throughout. Any horizontal line, regardless of its height, always has a derivative of zero at every point.

For a general exponential \(a^x\), the derivative is \(a^x \ln(a)\), so for \(a=2\) this gives \(2^x \ln(2)\). The choice \(2^x\) is wrong because it applies the rule for the special base \(e\), forgetting that other bases require multiplying by \(\ln(a)\). Only \(e^x\) has the unique property of being its own derivative; all other exponential bases require this extra logarithmic factor.

The function \(f(x)=x^3\) is a smooth polynomial with derivative \(3x^2\) defined everywhere including \(x=0\), giving \(f'(0)=0\) without any corner or cusp. The choice \(f(x)=|x|\) is wrong because it has a sharp corner at \(x=0\) where the left and right derivatives disagree, making it non-differentiable there. Polynomial functions are always differentiable everywhere, unlike absolute value or fractional power functions that can have cusps at specific points.