Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Chain Rule

Concept and Step-by-Step Application

• Definition. The chain rule states that if \( y = f(g(x)) \) is a composition of two functions, then \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x). \] This means you differentiate the outer function first (keeping the inside unchanged) and then multiply by the derivative of the inside. This rule accounts for how both the outer and inner functions change.
• Step-by-step process. 1) Identify the “outer” function and the “inner” function. 2) Differentiate the outer function with respect to the inner variable. 3) Multiply by the derivative of the inner function with respect to \( x \). 4) Simplify if needed. Example: \( \frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot 3 \).
• Why it works. The change in \( y \) is proportional to the change in the inner variable times the change in the outer variable. This relationship comes directly from the limit definition of the derivative and the chain of dependencies in composite functions.
• Example with multiple layers. For \( y = \sqrt{5x^2 + 1} \), the outer function is \( \sqrt{u} \) (or \( u^{1/2} \)) and the inner function is \( u = 5x^2 + 1 \). Differentiating: \( \frac{dy}{dx} = \frac{1}{2\sqrt{5x^2 + 1}} \cdot (10x) = \frac{10x}{2\sqrt{5x^2 + 1}} \).
• Common mistakes. Forgetting to multiply by the derivative of the inside function is the most common error. Another mistake is differentiating the inside first instead of the outer, which changes the meaning entirely. Always label “outer” and “inner” before starting.

Extended Chain Rule (Multiple Compositions)

• Formula. For \( y = f(g(h(x))) \), the derivative is \[ \frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x). \] This works the same way as the basic chain rule but continues through multiple layers.
• Step-by-step example. \( y = \sin^4(2x) \) means \( [\sin(2x)]^4 \). Outer function: \( u^4 \), derivative \( 4u^3 \). Inner function 1: \( \sin(2x) \), derivative \( \cos(2x) \). Inner function 2: \( 2x \), derivative \( 2 \). Multiply all: \( y' = 4\sin^3(2x) \cdot \cos(2x) \cdot 2 = 8\sin^3(2x)\cos(2x) \).
• Applications. Extended chain rule is essential for derivatives involving powers of trig, exponentials of logs, or radicals of polynomials. It appears in physics formulas where variables depend on intermediate quantities.
• Connections. The chain rule links directly to implicit differentiation (next section), because implicit differentiation is essentially applying the chain rule to \( y \) when it is treated as a function of \( x \).
• Common mistakes. Skipping an inner derivative in multi-layer problems leads to missing a factor, producing completely wrong answers. Also, misidentifying the true outermost layer can cause the wrong derivative structure.

Derivatives of Composite Functions

Trig, Exponential, Logarithmic, and Power Compositions

• Trig with inner functions. Example: \( \frac{d}{dx}[\sin(5x^2)] = \cos(5x^2) \cdot (10x) \). Identify “outer” as \( \sin u \), derivative \( \cos u \), and “inner” as \( 5x^2 \), derivative \( 10x \). The chain rule ensures we account for the rate of change of the inside angle.
• Exponential with inner functions. \( \frac{d}{dx}[e^{3x+1}] = e^{3x+1} \cdot 3 \). Outer: \( e^u \), derivative \( e^u \); inner: \( 3x+1 \), derivative \( 3 \). This pattern appears frequently in growth/decay models with shifts or scaling in the exponent.
• Logarithmic with inner functions. \( \frac{d}{dx}[\ln(7x^2+1)] = \frac{1}{7x^2+1} \cdot 14x \). Outer: \( \ln u \), derivative \( 1/u \); inner: \( 7x^2+1 \), derivative \( 14x \). This is common when differentiating composite log expressions.
• Power of an inner function. \( \frac{d}{dx}[(4x^3 - x)^{5}] = 5(4x^3 - x)^{4} \cdot (12x^2 - 1) \). Outer: \( u^5 \), derivative \( 5u^4 \); inner: \( 4x^3 - x \), derivative \( 12x^2 - 1 \).
• Common pitfalls. Failing to multiply by the inner derivative or attempting to simplify the inner before differentiating (which can lead to incorrect terms if you’re not careful). The chain rule works without needing to expand the inside unless it simplifies calculation.

Multiple-Inner Compositions

• Example with exponential and trig. \( y = e^{\sin(3x)} \) → Outer: \( e^u \), derivative \( e^u \); Inner 1: \( \sin(3x) \), derivative \( \cos(3x) \); Inner 2: \( 3x \), derivative \( 3 \). Result: \( y' = e^{\sin(3x)} \cdot \cos(3x) \cdot 3 \).
• Example with log of a power. \( y = \ln(\sqrt{2x+5}) \) → Outer: \( \ln u \), derivative \( 1/u \); Inner 1: \( \sqrt{2x+5} = (2x+5)^{1/2} \), derivative \( \frac{1}{2}(2x+5)^{-1/2} \cdot 2 \). Combine to get \( y' = \frac{1}{\sqrt{2x+5}} \cdot \frac{1}{2\sqrt{2x+5}} \cdot 2 = \frac{1}{2x+5} \).
• Connections. Many AP questions combine chain rule with product or quotient rules, so being fluent in combining them is essential. Example: differentiating \( x^2 e^{\sin x} \) uses both product and chain rules.
• Tip. Always explicitly identify the layers in your function. This helps prevent skipping derivatives when there are multiple inner functions.
• Common mistakes. Students sometimes “flatten” multiple layers into one and miss parts of the derivative. Writing each layer out in substitution form (e.g., \( u, v \)) before differentiating keeps the process organized.

Implicit Differentiation

Concept and Method

• When to use it. Implicit differentiation is used when \( y \) is defined implicitly in an equation with \( x \) rather than as an explicit \( y = f(x) \). This is common with circles, ellipses, and other curves that are not functions in the \( y = \) form.
• Basic idea. Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) and using the chain rule when differentiating terms with \( y \). Every time you differentiate \( y \), multiply by \( \frac{dy}{dx} \).
• Step-by-step example. Given \( x^2 + y^2 = 25 \): Differentiate → \( 2x + 2y\frac{dy}{dx} = 0 \). Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = -\frac{x}{y} \). This gives the slope of the tangent line at any point on the circle.
• Why it works. This method applies the chain rule to \( y \) because \( y \) depends on \( x \) even if we don’t know the exact formula. It allows slope calculation without explicitly solving for \( y \).
• Common mistakes. Forgetting the \( \frac{dy}{dx} \) factor when differentiating a \( y \)-term, or trying to solve for \( y \) before differentiating when it’s not feasible. Also, mixing up \( y \) and \( \frac{dy}{dx} \) during algebra simplification can lead to errors.

Applications and Extensions

• Slopes of tangent lines. Once \( \frac{dy}{dx} \) is found, substitute the coordinates of the point to get the slope of the tangent line at that point.
• Related rates problems. Implicit differentiation is essential in related rates because it allows you to differentiate with respect to time \( t \) when multiple variables depend on \( t \).
• Higher-order derivatives. After finding \( \frac{dy}{dx} \) implicitly, you can differentiate again (still implicitly) to find \( \frac{d^2y}{dx^2} \) for curvature or concavity analysis.
• Example. For \( xy = 1 \), differentiate: \( y + x\frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = -\frac{y}{x} \). Differentiating again: \( \frac{d^2y}{dx^2} = -\frac{dy/dx \cdot x - y}{x^2} \), then substitute \( \frac{dy}{dx} \) into the result.
• Connections. This method connects to the chain rule (every \( y \)-term is an inner function) and higher-order derivatives (where you repeatedly apply implicit differentiation).

Differentiation of Inverse Functions

General Formula and Why It Works

• Statement of the inverse-derivative formula. If \( f \) is differentiable with a nonzero derivative near \( a \), and \( f \) has an inverse \( f^{-1} \) near \( b=f(a) \), then \[ (f^{-1})'(b)=\frac{1}{f'(a)}=\frac{1}{f'\!\big(f^{-1}(b)\big)}. \] This expresses the slope of the inverse at an output \( b \) in terms of the slope of the original function at the corresponding input \( a \). The requirement \( f'(a)\neq 0 \) ensures local invertibility and avoids vertical tangents in the inverse.
• Reasoning via chain rule. The identity \( f\!\big(f^{-1}(x)\big)=x \) holds for \( x \) in the inverse’s domain. Differentiating both sides gives \( f'\!\big(f^{-1}(x)\big)\cdot (f^{-1})'(x)=1 \) by the chain rule. Solving for \( (f^{-1})'(x) \) yields \( \dfrac{1}{f'\!\big(f^{-1}(x)\big)} \), which is the general formula used in AP problems.
• Geometric interpretation. At a corresponding pair of points \( (a,f(a)) \) on \( y=f(x) \) and \( (f(a),a) \) on \( y=f^{-1}(x) \), the tangent slopes are reciprocal when neither slope is zero or infinite. A steep slope on \( f \) implies a shallow slope on \( f^{-1} \), reflecting how “stretching” in one direction becomes “compressing” in the inverse.
• Continuity and monotonicity conditions. For a local inverse to exist, \( f \) must be differentiable and have \( f'(a)\neq 0 \) (by the Inverse Function Theorem). Strict monotonicity on an interval guarantees a global inverse there, simplifying questions about domain and range for \( f^{-1} \).
• Common mistakes. Students often try to compute \( (f^{-1})'(x) \) by literally inverting the function first; this is unnecessary and frequently impossible in closed form. Others plug \( x=a \) into \( f' \) instead of \( x=f^{-1}(b) \); always evaluate \( f' \) at the matching input of \( f \), not at the inverse’s \( x \)-coordinate directly unless you convert it via \( a=f^{-1}(b) \).

Point-Based Evaluation and Table/Graph Problems

• Point method. To find \( (f^{-1})'(b) \) when given a point on \( f \), first locate \( a \) such that \( f(a)=b \). Then compute \( f'(a) \) (from a formula, graph slope, or table) and take its reciprocal: \( (f^{-1})'(b)=1/f'(a) \). This bypasses solving explicitly for \( f^{-1} \).
• Example (table). Suppose a table gives \( f(2)=7 \) and \( f'(2)=4 \). Then \( (f^{-1})'(7)=\dfrac{1}{f'(2)}=\dfrac{1}{4} \). If the table also lists \( f(3)=9 \), you can similarly conclude \( (f^{-1})'(9)=\dfrac{1}{f'(3)} \) once \( f'(3) \) is known.
• Example (graph). If the graph of \( y=f(x) \) shows a tangent at \( x=1 \) with slope \( -2 \) and passes through \( (1,5) \), then \( (f^{-1})'(5)=\dfrac{1}{-2}=-\dfrac{1}{2} \). Estimating slopes from the graph requires careful attention to scale and units on both axes.
• Units and interpretation. If \( f \) maps input units “seconds” to output units “meters,” then \( f'(a) \) has units m/s; its reciprocal \( (f^{-1})'(b) \) has units s/m. This helps catch sign and magnitude errors by checking whether the units make sense.
• Common pitfalls. Mixing up which value is the input for \( f' \) (must be \( a=f^{-1}(b) \)) and failing to verify \( f'(a)\neq 0 \). When a table is coarse, avoid assuming linear behavior across wide intervals; read or approximate the slope at the correct \( a \) as best as the data allow.

Derivatives of Inverse Trigonometric Functions

Core Formulas and Derivations (Domains Matter)

• Key results (for \( |x|<1 \) where required). \[ \frac{d}{dx}\big(\arcsin x\big)=\frac{1}{\sqrt{1-x^{2}}},\quad \frac{d}{dx}\big(\arccos x\big)=-\frac{1}{\sqrt{1-x^{2}}},\quad \frac{d}{dx}\big(\arctan x\big)=\frac{1}{1+x^{2}}. \] These come from implicit differentiation of \( \sin y=x, \cos y=x, \tan y=x \) with ranges chosen so the inverses are single-valued.
• Why \(\arcsin\) works. If \( y=\arcsin x \), then \( \sin y=x \) and \( \cos y\,\dfrac{dy}{dx}=1 \) by differentiating implicitly. Using \( \cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}} \) (with the principal branch), we get \( \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^{2}}} \). The square root reflects the restricted range for \( y \in [-\pi/2,\pi/2] \).
• Why \(\arccos\) gets a negative. If \( y=\arccos x \), then \( \cos y=x \) and \( -\sin y\,\dfrac{dy}{dx}=1 \). Because \( \sin y=\sqrt{1-\cos^{2}y}=\sqrt{1-x^{2}} \) on the principal range \( [0,\pi] \), we find \( \dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^{2}}} \). The negative is consistent with the decreasing nature of \( \arccos x \).
• \(\arctan\) derivation. If \( y=\arctan x \), then \( \tan y=x \) and \( \sec^{2}y\,\dfrac{dy}{dx}=1 \). Since \( \sec^{2}y=1+\tan^{2}y=1+x^{2} \), we get \( \dfrac{dy}{dx}=\dfrac{1}{1+x^{2}} \). This derivative is defined for all real \( x \) and aligns with the smooth, monotone growth of \( \arctan x \).
• Other inverse trig derivatives. You should know \[ \frac{d}{dx}(\mathrm{arccsc}\,x)=-\frac{1}{|x|\sqrt{x^{2}-1}},\quad \frac{d}{dx}(\mathrm{arcsec}\, x)=\frac{1}{|x|\sqrt{x^{2}-1}},\quad \frac{d}{dx}(\mathrm{arccot}\, x)=-\frac{1}{1+x^{2}}, \] with their appropriate domains. Absolute values reflect principal-branch choices and ensure correct signs where the functions are defined.

Applications, Signs, and Domain/Range Awareness

• Compositions via chain rule. For \( y=\arcsin(u(x)) \), use \( y'=\dfrac{u'(x)}{\sqrt{1-u(x)^{2}}} \) provided \( |u(x)|<1 \). Similarly, \( \dfrac{d}{dx}[\arctan(u(x))]=\dfrac{u'(x)}{1+u(x)^{2}} \). Always apply chain rule carefully to the inner function \( u(x) \).
• Example. Differentiate \( y=\arctan(3x^{2}) \): \( y'=\dfrac{6x}{1+9x^{4}} \). The denominator is always positive, so the sign of \( y' \) matches the sign of \( x \), reflecting the function’s increasing/decreasing behavior around 0.
• Domain and range checks. Inverse trig functions have restricted ranges, which affect sign choices when simplifying radicals (e.g., \( \cos(\arcsin x)=\sqrt{1-x^{2}} \) for \( x \in [-1,1] \)). Remember that \( \arccos \) is decreasing on \( [-1,1] \), so its derivative is negative.
• Units and modeling. When angles represent physical quantities, ensure you’re differentiating with respect to the correct variable (usually time or a linear dimension). The chain rule captures how changing inputs propagate through trig inverses to change angles.

Logarithmic Differentiation

Method and Step-by-Step Workflow

• When to use it. Apply logarithmic differentiation to functions that are products, quotients, or powers with variable exponents such as \( y=\big(f(x)\big)^{g(x)} \), or large products like \( y=\prod_{k} u_{k}(x) \). Taking logs turns products into sums and powers into multipliers, simplifying differentiation.
• Core idea. Start with \( y>0 \) and take natural logs: \( \ln y=\ln(\text{expression}) \). Differentiate both sides with respect to \( x \), using implicit differentiation on \( \ln y \) to get \( \dfrac{y'}{y} \). Solve for \( y' \) by multiplying both sides by \( y \) (i.e., the original function).
• Example (product/quotient/power mix). For \( y=\dfrac{(x^{2}+1)^{5}\sqrt{3x-1}}{e^{x}\sin^{2}x} \), write \[ \ln y=5\ln(x^{2}+1)+\tfrac{1}{2}\ln(3x-1)-x-2\ln(\sin x). \] Differentiate term-by-term, then multiply by \( y \) to get \( y' \). This avoids repeated product/quotient/chain-rule explosions.
• Example (variable exponent). For \( y=(x^{x}) \) on \( x>0 \), \( \ln y=\ln(x^{x})=x\ln x \Rightarrow \dfrac{y'}{y}=\ln x+1 \Rightarrow y'=x^{x}(\ln x+1) \). Log differentiation elegantly handles \( x^{x} \) without rewriting as \( e^{x\ln x} \) first (though that is equivalent).
• Common mistakes. Forgetting to multiply by \( y \) after differentiating \( \ln y \), dropping domain restrictions (e.g., \( 3x-1>0 \) for \( \ln(3x-1) \)), or mishandling the derivative of \( \ln(\sin x) \) which is \( \cot x \) times chain rule if there’s an inner function. Always restore \( y \) at the end using the original expression.

Applications and Efficiency Tips

• Large products and sums of logs. When \( y=\prod_{k=1}^{n} u_{k}(x) \), \( \ln y=\sum_{k=1}^{n}\ln u_{k}(x) \) turns one huge product into many manageable pieces. Differentiate term-by-term and reassemble to get \( y' \) succinctly.
• Precision and simplification. Log differentiation often yields cleaner algebra and reduces sign errors, especially when many chain, product, and quotient rules would otherwise stack. It is a preferred method on free-response when clarity matters.
• Connection to inverse functions. Since \( \ln \) is the inverse of \( e^{x} \), logarithmic differentiation leverages inverse properties to simplify exponentials and powers. Recognizing \( y=e^{\ln y} \) helps convert difficult forms into sums before differentiating.
• Combining with chain rule. Every \( \ln(u(x)) \) differentiates to \( \dfrac{u'(x)}{u(x)} \), which is a built-in chain rule. This consistency keeps the workflow uniform even for nested compositions.
• Quick check strategy. After obtaining \( y' \), do a units/behavior sanity check: if \( y \) increases rapidly (e.g., \( x^{x} \)), \( y' \) should also be large and positive for \( x>1 \). Check special points like \( x=1 \) where \( x^{x}=1 \) to verify \( y'=1\cdot(\ln 1+1)=1 \), which matches expectations.

Differentiation of General and Particular Inverse Functions

• To differentiate an inverse function \( f^{-1}(x) \), use the formula \( \frac{d}{dx} \big[ f^{-1}(x) \big] = \frac{1}{f'\big(f^{-1}(x)\big)} \). This works because inverse functions "undo" each other, so the chain rule applies in reverse. Always verify that \( f(x) \) is one-to-one and differentiable before applying this rule.
• Example: If \( f(x) = x^3 + 1 \), then \( f^{-1}(x) = \sqrt[3]{x-1} \). Differentiate by finding \( f'(x) = 3x^2 \), then \( \frac{d}{dx}f^{-1}(x) = \frac{1}{3\big( \sqrt[3]{x-1} \big)^2} \). This method avoids having to explicitly solve for \( f^{-1}(x) \) when it's hard to isolate.
• Particular inverse functions like \( \arcsin(x) \), \( \arccos(x) \), and \( \arctan(x) \) have standard derivatives: \( \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}} \), \( \frac{d}{dx}[\arccos(x)] = -\frac{1}{\sqrt{1-x^2}} \), and \( \frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2} \). These are derived from implicit differentiation of their trigonometric definitions.
• When differentiating inverse trig functions of more complex arguments (e.g., \( \arcsin(3x) \)), apply the chain rule after using the basic formula. For example: \( \frac{d}{dx}[\arcsin(3x)] = \frac{3}{\sqrt{1-9x^2}} \).
• Common mistake: forgetting to evaluate \( f' \) at \( f^{-1}(x) \) instead of \( x \) itself. This error leads to completely incorrect expressions, especially when the inverse is not a simple algebraic function.

Parametric Differentiation (Intro Level)

Definition and Purpose

• In parametric equations, both \( x \) and \( y \) are defined in terms of a third variable (the parameter), usually \( t \). For example: \( x(t) = t^2 + 1 \), \( y(t) = \sin t \). Parametric differentiation helps us find \( \frac{dy}{dx} \) without eliminating \( t \).
• The derivative \( \frac{dy}{dx} \) is found using the chain rule: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This works because \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) follows directly from dividing both rates of change with respect to \( t \).
• When \( \frac{dx}{dt} = 0 \), the tangent line is vertical because the change in \( x \) is zero at that moment. Similarly, if \( \frac{dy}{dt} = 0 \) but \( \frac{dx}{dt} \neq 0 \), the tangent line is horizontal.
• This method is especially useful in physics, motion problems, and calculus modeling, where eliminating \( t \) can be messy or impossible. It connects directly to earlier derivative rules and builds toward curvature and arc length later.
• Example: If \( x = t^2 + 2t \) and \( y = t^3 \), then \( \frac{dy}{dt} = 3t^2 \) and \( \frac{dx}{dt} = 2t + 2 \), so \( \frac{dy}{dx} = \frac{3t^2}{2t + 2} \).

Common Mistakes

• Forgetting to divide \( dy/dt \) by \( dx/dt \) and instead differentiating \( y \) with respect to \( x \) directly without parametric rules.
• Not checking for vertical or horizontal tangents by examining when \( dx/dt \) or \( dy/dt \) equals zero.
• Misapplying the quotient rule inside the parametric derivative — remember the division is only between the two derivatives, not the original equations.

Determining Higher-Order Derivatives of Functions

Definition and Process

• The first derivative \( f'(x) \) represents the rate of change of the function. The second derivative \( f''(x) \) represents the rate of change of the first derivative, telling us about concavity and acceleration. Higher-order derivatives are found by differentiating repeatedly.
• To find \( f''(x) \), take the derivative of \( f'(x) \). To find \( f'''(x) \), take the derivative of \( f''(x) \), and so on. For example, if \( f(x) = x^4 \), then \( f'(x) = 4x^3 \), \( f''(x) = 12x^2 \), \( f'''(x) = 24x \), and \( f^{(4)}(x) = 24 \).
• In parametric form, the second derivative requires an extra step: \( \frac{d^2y}{dx^2} = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} \). This formula ensures that both derivatives are properly expressed in terms of \( x \).
• Higher-order derivatives connect directly to Taylor polynomials, motion problems, and curve sketching. For example, in physics, \( y''(t) \) can represent acceleration, and \( y'''(t) \) can represent jerk (rate of change of acceleration).
• Example: If \( y = e^{2x} \), then \( y' = 2e^{2x} \), \( y'' = 4e^{2x} \), and \( y^{(n)} = 2^n e^{2x} \) by pattern recognition.

Step-by-Step Process

• Higher-order derivatives are obtained by repeatedly differentiating. The second derivative is \( \frac{d^2y}{dx^2} \), which measures the rate of change of the slope (concavity). For parametric equations: \[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}. \] This formula comes from applying the chain rule again, but now to \( \frac{dy}{dx} \) as a function of \( t \).
• Example: Continuing the example above where \( \frac{dy}{dx} = \frac{3t}{2} \): \[ \frac{d}{dt} \left( \frac{3t}{2} \right) = \frac{3}{2}, \quad \frac{dx}{dt} = 2t, \] so \[ \frac{d^2y}{dx^2} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t}. \] This tells us how the slope changes with \( x \).
• Reasoning: The second derivative is important for identifying concavity and inflection points, which play a key role in graph analysis and motion studies.
• Connection: In AP Calculus, higher-order derivatives are essential for curvature formulas, Taylor polynomials, and motion analysis (acceleration is the second derivative of position).
• Common Mistake: Mixing up the derivative with respect to \( t \) and with respect to \( x \), or forgetting to divide by \( dx/dt \) when finding \( d^2y/dx^2 \) in parametric form.

Parametric Differentiation (Intro Level)

Understanding the Basics

• In parametric equations, \( x \) and \( y \) are expressed as separate functions of a third variable, usually \( t \) (the parameter). The goal of parametric differentiation is to find \( \frac{dy}{dx} \) even though both \( x \) and \( y \) depend on \( t \). We use the chain rule in the form: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] This works because \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) links the rates of change with respect to \( t \).
• Example: If \( x = t^2 + 1 \) and \( y = t^3 \), then: \[ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 \] so \[ \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}. \] This tells us the slope at a given \( t \) without needing an explicit \( y(x) \) equation.
• Reasoning: We divide \( dy/dt \) by \( dx/dt \) because both derivatives measure change per unit of \( t \), so the ratio measures how much \( y \) changes for a given change in \( x \). This preserves the definition of slope while accommodating the parametric form.
• Connection: This method connects directly to chain rule ideas and will be critical when we study motion in the plane, where \( t \) often represents time and we want velocity, acceleration, and curvature.
• Common Mistake: Forgetting to compute \( dx/dt \) before dividing or not checking that \( dx/dt \neq 0 \). If \( dx/dt = 0 \), the slope is undefined (vertical tangent).