≡

Unit 2: Differentiation

Students will apply limits to define the derivative, become skillful at determining derivatives, and continue to develop mathematical reasoning skills.

Defining the Derivative of a Function at a Point

Limit Definition of the Derivative

  • The derivative of a function at a point \( x = a \) is defined as \[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \] provided the limit exists. This measures the slope of the secant line between \( x = a \) and \( x = a+h \) as \( h \) approaches zero, becoming the slope of the tangent line. If the limit fails to exist (e.g., due to discontinuity or sharp corners), the function is not differentiable at \( a \).
  • To evaluate this limit, you typically simplify the difference quotient algebraically, then substitute \( h = 0 \) after cancellation. Example: For \( f(x) = x^2 \), \[ f'(a) = \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} = \lim_{h \to 0} \frac{2ah + h^2}{h} = 2a \] which shows the slope depends linearly on \( a \).
  • Graphically, \( f'(a) \) is the slope of the tangent line to the curve at \( x=a \). This tangent line represents the instantaneous direction of the function’s growth or decay at that point, not the overall average slope over an interval.
  • Physically, the derivative at a point can represent instantaneous velocity (change of position with respect to time) or any other instantaneous rate of change, depending on the context. For example, if \( s(t) \) is position, \( s'(t) \) is velocity at time \( t \).
  • Common mistakes include trying to plug in \( h=0 \) too early (before simplification) and assuming a derivative exists just because the function is defined at \( a \). The limit must exist and be finite from both sides for differentiability.

Alternate Forms of the Definition

  • Another form of the derivative definition uses \( x \) approaching \( a \): \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} \] This form is especially useful when given a limit expression directly in terms of \( x \) and \( a \), such as in AP free-response questions.
  • Both forms are equivalent; the \( h \)-form emphasizes change from a fixed point, while the \( x \)-form emphasizes approaching \( a \) directly. Knowing both allows flexibility in solving problems.
  • Example: Using \( x \)-form for \( f(x) = \sqrt{x} \) at \( a = 4 \): \[ f'(4) = \lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{4} \]
  • Both definitions require the function to be defined near \( a \) and to have a single slope value from both sides. If the left and right slopes differ, the derivative does not exist at that point.
  • Common pitfalls include forgetting to rationalize when square roots are present, or misinterpreting the limit’s result as an average rate instead of an instantaneous rate.

Defining the Derivative as a Function

Derivative Function Concept

  • The derivative function \( f'(x) \) gives the slope of the tangent line to \( f \) at any point \( x \). It is found by applying the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] and leaving \( x \) as a variable instead of plugging in a specific value.
  • Example: For \( f(x) = x^2 \), \[ f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = 2x \] This shows the derivative function produces the slope for any \( x \), not just one point.
  • The derivative function is itself a function and can be graphed, analyzed, and differentiated again. Its sign tells us where \( f \) is increasing (\( f'(x) > 0 \)) or decreasing (\( f'(x) < 0 \)).
  • In applied contexts, \( f'(x) \) can model changing rates over time, such as acceleration when \( f(x) \) is velocity. This connects directly to higher-order derivatives later in the course.
  • Common errors include treating \( f'(x) \) as a single number, forgetting it varies with \( x \), and mixing up the roles of \( x \) and \( h \) when using the limit definition.

Numerical and Graphical Estimation

  • The derivative function can be estimated numerically using difference quotients with small \( h \) values for various \( x \). This approach is useful when an explicit formula for \( f \) is unknown but data points are given.
  • Graphically, the slope of the tangent line at various \( x \)-values can be approximated and plotted to visualize \( f'(x) \). This is especially important in interpreting motion graphs.
  • Example: If \( f(t) \) represents position at time \( t \), plotting the slope of tangents on \( f(t) \) vs. \( t \) produces the velocity-time graph, which can then be analyzed for acceleration.
  • This connects directly to calculus concepts in later units, such as optimization and motion analysis, where understanding how \( f'(x) \) behaves is essential.
  • Common mistakes include misreading slopes on a graph (especially if axes are scaled differently) and assuming the derivative’s magnitude alone determines the function’s shape without considering sign.

Connecting Differentiability and Continuity

Relationship Between the Two

  • If a function is differentiable at \( x = a \), then it must be continuous at \( x = a \). Differentiability is a stronger condition: it requires not only that \( f(a) \) match the limit from both sides, but also that the slopes from both sides are equal and finite.
  • Continuity does not imply differentiability. Functions can be continuous but not differentiable at points where there are sharp corners (e.g., \( |x| \) at 0), cusps (e.g., \( x^{2/3} \) at 0), or vertical tangents (e.g., \( x^{1/3} \) at 0).
  • Example: \( f(x) = |x| \) is continuous everywhere but has no derivative at \( x = 0 \) because the left-hand slope is -1 and the right-hand slope is +1. This mismatch means the derivative limit does not exist.
  • In motion problems, non-differentiable points correspond to abrupt changes in velocity, which would require infinite acceleration — physically unrealistic in many contexts but possible in idealized models.
  • Common misconceptions include thinking differentiability can “fix” discontinuity or assuming a piecewise function is differentiable at its joint just because it’s continuous there. Always check the slope from both sides to confirm.

Testing for Differentiability

  • To test if \( f \) is differentiable at \( x = a \): (1) Check if \( f \) is continuous at \( a \). (2) Compute \( f'(a^-) \) and \( f'(a^+) \) from the left and right. (3) Confirm both one-sided derivatives exist and are equal.
  • Example: For \( f(x) = \begin{cases} x^2, & x<1 \\ 2x-1, & x\ge1 \end{cases} \), the function is continuous at \( x=1 \) (both sides give 1). Left-hand derivative: \( 2x \) at \( x=1 \) is 2; right-hand derivative: derivative of \( 2x-1 \) is 2. Equal slopes mean differentiable at 1.
  • Vertical tangents occur when \( f'(a^-) \) and \( f'(a^+) \) both approach \( \pm\infty \) but match in sign. This means the function is not differentiable at that point despite continuous shape.
  • This topic directly links to Unit 1’s continuity and one-sided limits, reinforcing that derivative existence depends on limit behavior from both directions.
  • Common mistakes include skipping the slope equality check and only verifying continuity, or miscomputing one-sided derivatives for piecewise definitions.

Determining Derivatives for Elementary Functions

Power and Constant Functions

  • Power rule statement and meaning. For any real exponent \( n \), the derivative of \( f(x)=x^{n} \) is \( f'(x)=n x^{n-1} \) wherever the expression makes sense. This comes from applying the limit definition to the binomial/series expansion and seeing the linear term dominate as \( h\to 0 \). Conceptually, it tells you how the slope scales with both the exponent and the current input value.
  • How to apply it step-by-step (example 1). Differentiate \( f(x)=x^{5} \): bring down the exponent and subtract one to get \( f'(x)=5x^{4} \). This requires no product or chain rule because it is a single power of \( x \). Always check whether the base is just \( x \) or a composition like \( (3x-1)^{5} \) (which needs chain rule later).
  • Constants and constant multiples. If \( f(x)=C \) (a horizontal line), then \( f'(x)=0 \) because the slope is zero everywhere. If \( f(x)=C\cdot g(x) \), then \( f'(x)=C\cdot g'(x) \), meaning constants “pass through” the derivative operator. This simplifies long expressions by letting you factor constants out before differentiating.
  • Negative and fractional powers. The power rule still applies: e.g., \( \dfrac{d}{dx}\big(x^{-2}\big)=-2x^{-3} \) and \( \dfrac{d}{dx}\big(x^{1/2}\big)=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}} \) for \( x>0 \). Be mindful of domains because fractional powers may restrict \( x \) and negative powers introduce vertical asymptotes where \( x=0 \). Domain awareness connects to Unit 1’s continuity discussion.
  • Common mistakes and quick checks. Students often forget to decrement the exponent or they treat \( (ax)^{n} \) as \( a x^{n} \) without accounting for \( a^{n} \). Confirm the base is exactly \( x \); if it is \( (ax+b)^{n} \), you will need chain rule (covered later). Sanity-check by estimating slope at a simple point, e.g., for \( x^{2} \) at \( x=1 \), \( f'(1)=2 \) matches the graph’s slope.

Trigonometric, Exponential, and Logarithmic Functions

  • Core trig derivatives (radians required). Memorize \( (\sin x)'=\cos x \), \( (\cos x)'=-\sin x \), and \( (\tan x)'=\sec^{2}x \), with \( (\csc x)'=-\csc x\cot x \), \( (\sec x)'=\sec x\tan x \), \( (\cot x)'=-\csc^{2}x \). These follow from the special trig limits in Unit 1 and are valid only in radian measure. Always confirm the problem is in radians on AP exams.
  • Exponential functions. \( (e^{x})'=e^{x} \) because \( e^{x} \) is its own rate of change; for general bases \( a>0 \), \( (a^{x})'=a^{x}\ln a \). Example: \( \dfrac{d}{dx}\big(3\cdot 2^{x}\big)=3\cdot 2^{x}\ln 2 \). Recognize that the natural base \( e \) simplifies work and appears often in modeling.
  • Logarithms. \( (\ln x)'=\dfrac{1}{x} \) for \( x>0 \) and \( (\log_{a} x)'=\dfrac{1}{x\ln a} \). Example: \( \dfrac{d}{dx}\big(\ln(5x)\big)=\dfrac{1}{5x}\cdot 5=\dfrac{1}{x} \) if treated by chain rule (formally covered later). Logs convert multiplicative changes into additive rates, which helps in growth/decay problems.
  • Worked example combining types. Differentiate \( f(x)=x^{2}\sin x+\ln x \) (for \( x>0 \)): the derivative is \( f'(x)=2x\sin x+x^{2}\cos x+\dfrac{1}{x} \) using sum, product (for \( x^{2}\sin x \)), and log rules. Each term’s rule is applied locally, then combined into a single expression.
  • Common mistakes and connections. Do not use degree mode; derivatives of trig functions assume radians. Do not confuse \( (\ln x)' \) with \( \ln(x)' \) notation; the derivative is \( 1/x \), not “\( \ln \) prime.” These elementary derivatives are the building blocks for the product, quotient, and chain rules in later subtopics.

Basic Differentiation Rules

Constant, Constant Multiple, and Sum/Difference Rules

  • Linearity of the derivative. The derivative is linear: \( (c)'\!=0 \), \( (c\,f)'\!=c\,f' \), and \( (f\pm g)'\!=f'\pm g' \). Linearity allows term-by-term differentiation, dramatically simplifying long expressions. This is especially helpful before applying product/quotient/chain rules.
  • Step-by-step example (polynomial). For \( p(x)=4x^{5}-3x^{2}+7 \), apply term-wise rules: \( p'(x)=4\cdot 5x^{4}-3\cdot 2x+0=20x^{4}-6x \). Notice constants vanish and coefficients pass through unchanged. This pattern generalizes to any polynomial.
  • Combining with elementary derivatives. If \( F(x)=3e^{x}-2\sin x+\ln x \) (domain \( x>0 \)), then \( F'(x)=3e^{x}-2\cos x+\dfrac{1}{x} \). Each term uses its own elementary rule, then results are added. Always keep domains in mind when logs are present.
  • Strategic factoring before differentiating. Sometimes factor a constant or simple expression to expose a cleaner structure. For example, \( \dfrac{d}{dx}[5(x^{3}-x)]=5(3x^{2}-1) \). This avoids reapplying product rule unnecessarily when factors are constants.
  • Common mistakes and checks. Do not distribute derivatives over products (that is a different rule). Watch signs when differentiating differences, and treat each term independently. A quick check is to test a simple point numerically to ensure the slope’s sign and magnitude are reasonable.

Power Rule and Notation (Prime, Leibniz, and Differential Form)

  • Notation equivalence. \( f'(x) \), \( \dfrac{dy}{dx} \), and \( D_{x}[f(x)] \) all denote the derivative of \( y=f(x) \) with respect to \( x \). Higher derivatives use \( f''(x) \), \( \dfrac{d^{2}y}{dx^{2}} \), etc. Choose the notation that best fits the problem’s context (graphical vs. algebraic vs. differential equations).
  • Applying the power rule in each notation. If \( y=x^{n} \), then \( \dfrac{dy}{dx}=n x^{n-1} \) and \( y'=n x^{n-1} \). Practice translating between notations seamlessly so you can follow AP FRQ prompts regardless of their style.
  • Differentials perspective. Writing \( dy=f'(x)\,dx \) interprets the derivative as a local linear scale factor between changes in \( x \) and changes in \( y \). This viewpoint becomes useful for error propagation and differential approximations in later units.
  • Worked example with mixed notation. Given \( y=7x^{-3/2} \), \( \dfrac{dy}{dx}=7\left(-\dfrac{3}{2}\right)x^{-5/2}=-\dfrac{21}{2}x^{-5/2} \). Rewriting with radicals: \( y'=-\dfrac{21}{2\;x^{5/2}}=-\dfrac{21}{2x^{2}\sqrt{x}} \) for \( x>0 \). Notation changes do not alter the calculus.
  • Common pitfalls. Students sometimes treat \( \dfrac{dy}{dx} \) as a fraction in all settings; while it behaves fraction-like in differential manipulations, remember it is a limit. Also, confusing \( d^{2}y/dx^{2} \) with \( (dy/dx)^{2} \) is a frequent error; the former is the second derivative, not a square.

Applying Differentiation Rules

Polynomials and Sums: Step-by-Step Workflow

  • Plan the attack. Identify term types (powers, exponentials, trig, logs) and mark where product/quotient/chain rules will be needed. Factor out constants first and reorder terms if it reveals simpler patterns. A clear plan reduces sign mistakes and missed rules.
  • Worked example (mixed polynomial/trig/log). For \( f(x)=2x^{4}-3x\cos x+\ln(2x) \), handle each term: \( (2x^{4})'=8x^{3} \); \( (-3x\cos x)'=-3(\cos x - x\sin x) \) by product rule; \( (\ln(2x))'=\dfrac{1}{2x}\cdot 2=\dfrac{1}{x} \) by chain rule with \( u=2x \). Combine to get \( f'(x)=8x^{3}-3\cos x+3x\sin x+\dfrac{1}{x} \) for \( x>0 \).
  • Check domains and simplify. After differentiating, state domain restrictions (e.g., \( x>0 \) for logs, \( x\neq 0 \) for negative powers) and simplify algebraically only when it improves clarity. Domain statements connect back to continuity and differentiability from Unit 1.
  • Interpret the result. Use signs of \( f'(x) \) to anticipate where the original function increases or decreases, previewing Unit 3 analysis. For the example above, \( 8x^{3} \) dominates for large \( |x| \), suggesting increasing end behavior.
  • Common mistakes and quick audit. Missing a product rule is the number-one error when a factor like \( x\cos x \) appears. Another is forgetting the inner derivative when differentiating \( \ln(ax) \) or \( a^{x} \). Audit your answer by scanning each original term and confirming an appropriate rule was applied once and only once.

Chain Rule Preview: Basic Compositions (No Full Generality Yet)

  • Idea of composition. When the input to a basic function is itself a function, you must account for how fast the inner input changes. The chain rule states \( \dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x) \). We preview simple cases to avoid misapplying the power rule.
  • Worked example (power of a linear function). \( \dfrac{d}{dx}\big((3x-1)^{5}\big)=5(3x-1)^{4}\cdot 3=15(3x-1)^{4} \). The extra factor \( 3 \) comes from the derivative of the inside \( g(x)=3x-1 \). Forgetting this factor is the classic chain-rule error.
  • Log and exponential with linear inner functions. \( \dfrac{d}{dx}\big(\ln(5x-2)\big)=\dfrac{1}{5x-2}\cdot 5=\dfrac{5}{5x-2} \) (domain \( 5x-2>0 \)); \( \dfrac{d}{dx}\big(e^{2x}\big)=e^{2x}\cdot 2 \). Recognize the pattern “outer derivative evaluated at inner” times “inner derivative.”
  • Trig with linear inner functions. \( \dfrac{d}{dx}\big(\sin(4x)\big)=\cos(4x)\cdot 4 \) and \( \dfrac{d}{dx}\big(\cos(2x-3)\big)=-\sin(2x-3)\cdot 2 \). This extends the elementary trig derivatives to more realistic inputs that appear in modeling.
  • Common pitfalls and connections. Misidentifying a composition as a simple power leads to missing the inner derivative. Always label “outer” and “inner” before differentiating; this habit will pay off when we formalize the chain rule and related techniques (implicit differentiation, inverse functions) next.

Product and Quotient Rules

Product Rule

  • Formula and concept. The product rule states that if \( h(x) = f(x) \cdot g(x) \), then \[ h'(x) = f'(x)g(x) + f(x)g'(x). \] This accounts for both functions changing simultaneously; you cannot simply differentiate each and multiply the results.
  • Step-by-step process. 1) Label \( f(x) \) and \( g(x) \). 2) Differentiate \( f(x) \) while keeping \( g(x) \) unchanged, then add \( f(x) \) unchanged times the derivative of \( g(x) \). 3) Simplify only if it improves clarity. Example: \( \dfrac{d}{dx} \left[ x^2\sin x \right] = (2x)\sin x + (x^2)(\cos x) \).
  • Why it works. The rule comes directly from the limit definition: the change in a product comes from the change in the first factor plus the change in the second factor, with both changes considered in small increments.
  • Common mistakes. Students often forget one of the terms, reverse the order of multiplication without regard to function type (especially for vector/matrix calculus later), or think they can multiply derivatives directly. Always ensure both terms are present.
  • Connections. The product rule is critical for differentiating physical formulas (e.g., \( F=ma(t) \), where both mass and acceleration vary) and appears in applications involving related rates and integration by parts.

Quotient Rule

  • Formula and concept. The quotient rule states that if \( h(x) = \dfrac{f(x)}{g(x)} \) and \( g(x) \neq 0 \), then \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{\left[g(x)\right]^2}. \] The order of subtraction matters: “low d-high minus high d-low, over low squared.”
  • Step-by-step process. 1) Identify numerator \( f(x) \) and denominator \( g(x) \). 2) Differentiate \( f(x) \) and multiply by \( g(x) \). 3) Subtract \( f(x) \) times the derivative of \( g(x) \). 4) Place the result over \( g(x)^2 \). Example: \( \dfrac{d}{dx} \left[ \dfrac{\sin x}{x} \right] = \dfrac{\cos x \cdot x - \sin x \cdot 1}{x^2} \).
  • Why it works. The quotient rule is derived from the product rule and chain rule applied to \( f(x) \cdot [g(x)]^{-1} \). Recognizing this connection helps in memorization and understanding.
  • Common mistakes. Forgetting to square the denominator, reversing the subtraction order (leading to an overall sign error), and neglecting to check domain restrictions where \( g(x) = 0 \) are typical errors.
  • Connections. The quotient rule is essential for differentiating rational functions, certain trigonometric expressions, and physical laws expressed as ratios (e.g., resistance, density, average rate formulas).

Higher-Order Derivatives

Definition and Notation

  • What they are. Higher-order derivatives are obtained by differentiating a derivative. The second derivative is \( f''(x) \) or \( \dfrac{d^2y}{dx^2} \), the third is \( f'''(x) \), and so on. This process can be repeated as long as the resulting function is differentiable.
  • Notation styles. Prime notation \( f''(x) \) is common for low orders; Leibniz notation \( \dfrac{d^n y}{dx^n} \) is preferred for higher orders. In applied contexts, dots over variables (Newton notation) may be used for derivatives with respect to time.
  • Step-by-step example. If \( f(x) = x^4 \), then \( f'(x) = 4x^3 \), \( f''(x) = 12x^2 \), \( f'''(x) = 24x \), and \( f^{(4)}(x) = 24 \). Each differentiation applies the rules from earlier sections independently.
  • Common mistakes. Forgetting that each differentiation starts fresh from the previous derivative (not the original function) and carrying forward algebra errors from earlier steps. Also, some confuse \( (f'(x))^2 \) with \( f''(x) \), which are completely different.
  • Connections. Second derivatives appear in concavity tests and acceleration in physics; third derivatives and beyond can model rates of change of rates (jerk in motion). Mastering basic differentiation rules ensures accuracy in higher orders.

Applications and Interpretation

  • In physics. First derivative of position with respect to time is velocity, second is acceleration, third is jerk. Each step reveals a deeper level of how the system changes over time.
  • In math analysis. Second derivatives test concavity and find points of inflection; higher derivatives are used in Taylor polynomial expansions and series approximations in AP Calculus BC’s later units.
  • Worked example (concavity). Given \( f(x) = x^3 - 3x^2 + 2 \), \( f''(x) = 6x - 6 \). Setting \( f''(x) = 0 \) gives \( x=1 \) as a possible inflection point; testing intervals shows concave down for \( x<1 \) and concave up for \( x>1 \).
  • Checking with graphs. Plotting the original function and its derivatives together clarifies the relationship: where \( f'(x) \) is increasing, \( f''(x) > 0 \) (concave up); where \( f'(x) \) is decreasing, \( f''(x) < 0 \) (concave down).
  • Common pitfalls. Forgetting that points where \( f''(x) = 0 \) may not be inflection points — you must check for a sign change. Also, for motion problems, mixing up position, velocity, and acceleration graphs is a frequent error.

Practice Problems

Problem 1 (Enforcing Differentiability at a Junction (Solve for Parameters))

  • Question. Let \[ f(x)= \begin{cases} x^{2}\sin x, & x<0,\\[4pt] a\,x+b+e^{x}, & x\ge 0. \end{cases} \] (a) Choose constants \( a \) and \( b \) so that \( f \) is continuous and differentiable at \( x=0 \). (b) Find the equation of the tangent line to \( f \) at \( x=0 \). Explain each step and why it works.
  • Step 1: Enforce continuity at \( x=0 \) (Unit 1 connection). Continuity requires \( \lim_{x\to 0^-}f(x)=f(0)=\lim_{x\to 0^+}f(x) \). Left side: \( \lim_{x\to 0^-} x^{2}\sin x = 0 \) (since \( x^{2}\to 0 \) and \( |\sin x|\le 1 \)). Right side: \( f(0)=a\cdot 0 + b + e^{0}=b+1 \). Set equal to 0 to get \( b+1=0 \Rightarrow b=-1 \). This matches the three-part continuity test from Unit 1.
  • Step 2: Enforce differentiability at \( x=0 \). Differentiability requires equal one-sided derivatives. Left derivative: \( \dfrac{d}{dx}(x^{2}\sin x)=2x\sin x + x^{2}\cos x \), so \( f'_-(0)=0 \). Right derivative: \( \dfrac{d}{dx}(a\,x+b+e^{x})=a+e^{x} \), hence \( f'_+(0)=a+1 \). Set \( f'_-(0)=f'_+(0) \Rightarrow 0=a+1 \Rightarrow a=-1 \). Differentiability ⇒ continuity, and we already satisfied continuity with \( b=-1 \).
  • Step 3: State the parameters and justify. The only choice that makes \( f \) continuous and differentiable at \( 0 \) is \( a=-1 \) and \( b=-1 \). Any other pair would either break continuity (value mismatch) or produce unequal one-sided derivatives (slope mismatch).
  • Step 4: Tangent line at \( x=0 \). With \( a=-1 \), \( b=-1 \): \( f(0)=0 \) and \( f'(0)=0 \). The tangent line is \( y=f(0)+f'(0)(x-0)=0 \), i.e., the \( x \)-axis. This follows from the point-slope form of a tangent using the derivative as slope.
  • Common mistakes & connections. A typical error is to enforce only continuity (\( b=-1 \)) and forget to match derivatives; another is to treat \( x^{2}\sin x \) as non-differentiable at 0 despite being a product of smooth functions. This problem explicitly connects Unit 1 (continuity, one-sided limits) with Unit 2 (derivative as a slope and the requirement \( f'_-(0)=f'_+(0) \)).

Problem 2 (Product & Quotient Rules with Higher-Order Derivatives)

  • Question. Define \[ h(x)=\frac{(x^{2}+1)\,e^{3x}}{x-1}. \] (a) Compute \( h'(x) \) using product/quotient rules. (b) Compute \( h''(0) \). Show a clean workflow and justify each rule you use.
  • Step 1: Organize with \( u\!/\!v \) structure. Let \( u(x)=(x^{2}+1)e^{3x} \) and \( v(x)=x-1 \), so \( h(x)=u/v \). The quotient rule gives \[ h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^{2}}. \] We will find \( u'(x) \) by the product rule since \( u=(x^{2}+1)\cdot e^{3x} \).
  • Step 2: Differentiate \( u \) via product rule. With \( p(x)=x^{2}+1 \) and \( q(x)=e^{3x} \), we have \( p'(x)=2x \) and \( q'(x)=3e^{3x} \). Thus \[ u'(x)=p'(x)q(x)+p(x)q'(x)=2x\,e^{3x}+(x^{2}+1)\cdot 3e^{3x} =e^{3x}\big(2x+3x^{2}+3\big). \] Also \( v'(x)=1 \).
  • Step 3: Assemble \( h'(x) \) and optionally factor. Substitute into the quotient rule: \[ h'(x)=\frac{\;e^{3x}(2x+3x^{2}+3)\,(x-1)\;-\;(x^{2}+1)e^{3x}\cdot 1\;}{(x-1)^{2}} =\frac{e^{3x}\Big[(2x+3x^{2}+3)(x-1)-(x^{2}+1)\Big]}{(x-1)^{2}}. \] Keeping \( e^{3x} \) factored helps later evaluations and reduces algebra slip-ups.
  • Step 4: Prepare for \( h''(0) \) efficiently. It’s faster to evaluate needed pieces at \( x=0 \) rather than fully expanding everything. Compute base values: \( e^{0}=1 \), \( (x-1)|_{0}=-1 \), \( (x^{2}+1)|_{0}=1 \), and from Step 2 \( u'(0)=e^{0}(0+0+3)=3 \); also \( u(0)=1 \), \( v(0)=-1 \), \( v'(0)=1 \).
  • Step 5: First derivative at 0 (as a checkpoint). Using \( h'=\dfrac{u'v-uv'}{v^{2}} \), \[ h'(0)=\frac{(3)(-1)-(1)(1)}{(-1)^{2}}=\frac{-3-1}{1}=-4. \] This gives a reliable intermediate check before differentiating again.
  • Step 6: Differentiate \( h' \) to get \( h'' \) (quotient rule again or use logarithmic differentiation idea). Continue with quotient rule on \( h'=\dfrac{N}{D} \) where \( N=u'v-uv' \) and \( D=v^{2} \). Then \[ h''(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^{2}}. \] At \( x=0 \), compute the needed pieces: • \( D(0)=v(0)^{2}=(-1)^{2}=1 \), \( D'(x)=2v(x)v'(x)\Rightarrow D'(0)=2(-1)(1)=-2 \). • \( N(x)=u'v-uv' \Rightarrow N(0)=3(-1)-1(1)=-4 \) (already used). • \( N'(x)=u''v+u'v'-u'v'-uv''=u''v-uv'' \) since the middle terms cancel. Here \( v''(x)=0 \), so \( N'(0)=u''(0)\,v(0) \).
  • Step 7: Compute \( u''(0) \) cleanly. From Step 2, \( u'(x)=e^{3x}(2x+3x^{2}+3) \). Differentiate: \[ u''(x)=3e^{3x}(2x+3x^{2}+3)+e^{3x}(2+6x) =e^{3x}\big(6x+9x^{2}+9+2+6x\big) =e^{3x}\big(9x^{2}+12x+11\big). \] Evaluate at 0: \( u''(0)=1\cdot 11=11 \). Thus \( N'(0)=u''(0)\,v(0)=11\cdot(-1)=-11 \).
  • Step 8: Finish \( h''(0) \). Substitute into the quotient-once-more formula at \( x=0 \): \[ h''(0)=\frac{N'(0)D(0)-N(0)D'(0)}{[D(0)]^{2}} =\frac{(-11)(1)-(-4)(-2)}{1} =-11-8=-19. \] Therefore \( h''(0)=-19 \). This uses a structured quotient-rule framework and targeted point evaluations to avoid unnecessary algebra.
  • Common mistakes & connections. Frequent errors include forgetting the outer \( (x-1)^{2} \) in denominators, dropping the product-rule term when differentiating \( u \), or expanding everything before substituting \( x=0 \) (which increases algebra risk). This problem synthesizes: product rule, quotient rule, exponential derivatives, and higher-order derivatives with an efficient evaluation strategy.