Trigonometric Limits, Chain Rule, Exponential Differentiation, and Implicit Differentiation

Trigonometric Limits and Simple Lemma

  • The core limit lemma for trigonometric simplification is given by:   lim⁡θ→0sin⁡(θ)θ=1\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1

  • Procedure for evaluating limits of the form lim⁡x→0sin⁡(kx)x\lim_{x \to 0} \frac{\sin(kx)}{x}:

    • Directly substituting x=0x = 0 yields an indeterminate form 00\frac{0}{0}.

    • To use the lemma, adjust the expression so the argument of the sine function matches the denominator exactly.

    • Multiply both the numerator and denominator by the constant factor kk:     lim⁡x→0sin⁡(kx)x=lim⁡x→0k×sin⁡(kx)kx=k×1=k\lim_{x \to 0} \frac{\sin(kx)}{x} = \lim_{x \to 0} k \times \frac{\sin(kx)}{kx} = k \times 1 = k

    • Example: For lim⁡x→0sin⁡(3x)x\lim_{x \to 0} \frac{\sin(3x)}{x}, multiply the numerator and denominator by 33:     lim⁡x→03×sin⁡(3x)3x=3×1=3\lim_{x \to 0} 3 \times \frac{\sin(3x)}{3x} = 3 \times 1 = 3

  • Procedure for evaluating limits involving tangent functions:

    • Rewrite tan⁡(θ)\tan(\theta) as sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)}.

    • For the limit lim⁡x→0tan⁡(5x)2x\lim_{x \to 0} \frac{\tan(5x)}{2x}:

    • Rewrite the expression as:       lim⁡x→0sin⁡(5x)2xcos⁡(5x)=lim⁡x→0(sin⁡(5x)x)×(12cos⁡(5x))\lim_{x \to 0} \frac{\sin(5x)}{2x \cos(5x)} = \lim_{x \to 0} \left(\frac{\sin(5x)}{x}\right) \times \left(\frac{1}{2\cos(5x)}\right)

    • Isolate the constant factor of 22 in the denominator to avoid introducing unnecessary multipliers.

    • Multiply and divide the fraction sin⁡(5x)x\frac{\sin(5x)}{x} by 55:       lim⁡x→0(5×sin⁡(5x)5x)×(12cos⁡(5x))\lim_{x \to 0} \left(5 \times \frac{\sin(5x)}{5x}\right) \times \left(\frac{1}{2\cos(5x)}\right)

    • Evaluate the limit as x→0x \to 0:       5×1×12cos⁡(0)=5×1×12(1)=525 \times 1 \times \frac{1}{2\cos(0)} = 5 \times 1 \times \frac{1}{2(1)} = \frac{5}{2}

The Chain Rule for Composite Functions

  • Definition and Formula:

    • The chain rule is used to differentiate composite functions f(g(x))f(g(x)).

    • Mathematical formula:     ddx[f(g(x))]=f′(g(x))×g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)

    • Differentiate the outer function with respect to the inner function (keeping the base unchanged), then multiply by the derivative of the inner function (the base) with respect to xx.

  • Relationship to standard power rule:

    • For a simple function P(x)=x7P(x) = x^7, differentiating gives P′(x)=7x6×dxdx=7x6×1=7x6P'(x) = 7x^6 \times \frac{dx}{dx} = 7x^6 \times 1 = 7x^6.

    • The chain rule is implicitly applied even in basic power rule differentiations, but the derivative of the base xx with respect to xx is 11, so it is omitted.

  • Direct differentiation example:

    • Given H(x)=(5x2+2)7H(x) = (5x^2 + 2)^7:

    • Identify the base as (5x2+2)(5x^2 + 2) and the exponent as 77.

    • Apply the outer power rule: 7(5x^2 + 2)^6$.\n - Multiply by the derivative of the base: \frac{d}{dx}(5x^2 + 2) = 10x + 0 = 10x.\n - Combine and simplify:\n      H'(x) = 7(5x^2 + 2)^6 \times 10x = 70x(5x^2 + 2)^6\n\n- Derivative using u-substitution (Leibniz notation):\n - Let u = 5x^2 + 2,so, soH(u) = u^7$.

    • According to the chain rule formula in Leibniz notation:     dHdx=dHdu×dudx\frac{dH}{dx} = \frac{dH}{du} \times \frac{du}{dx}

    • Compute each derivative separately:     dHdu=7u6\frac{dH}{du} = 7u^6     dudx=10x\frac{du}{dx} = 10x

    • Substitute back into the chain rule formula:     dHdx=(7u6)(10x)=7(5x2+2)6(10x)=70x(5x2+2)6\frac{dH}{dx} = (7u^6)(10x) = 7(5x^2 + 2)^6(10x) = 70x(5x^2 + 2)^6

Applications of the Chain Rule: Damped Harmonic Motion

  • Physical context:

    • A dashpot spring system subject to a damping force has a position function given by:     F(t)=e−2tsin⁡(3t)F(t) = e^{-2t} \sin(3t)

    • Velocity V(t)V(t) is the first derivative of position with respect to time tt:     V(t)=F′(t)V(t) = F'(t)

  • Differentiation process using Product Rule and Chain Rule:

    • Apply the product rule ddt[u⋅v]=u′v+uv′\frac{d}{dt}[u \cdot v] = u'v + uv' where u=e−2tu = e^{-2t} and v=sin⁡(3t)v = \sin(3t).

    • Differentiate u=e−2tu = e^{-2t} using the chain rule:     ddt(e−2t)=e−2t×ddt(−2t)=−2e−2t\frac{d}{dt}(e^{-2t}) = e^{-2t} \times \frac{d}{dt}(-2t) = -2e^{-2t}

    • Differentiate v=sin⁡(3t)v = \sin(3t) using the chain rule:     ddt(sin⁡(3t))=cos⁡(3t)×ddt(3t)=3cos⁡(3t)\frac{d}{dt}(\sin(3t)) = \cos(3t) \times \frac{d}{dt}(3t) = 3\cos(3t)

    • Combine using the product rule:     F′(t)=(−2e−2t)sin⁡(3t)+e−2t(3cos⁡(3t))F'(t) = (-2e^{-2t})\sin(3t) + e^{-2t}(3\cos(3t))

    • Factoring e−2te^{-2t} yields the simplified velocity function:     V(t)=e−2t(−2sin⁡(3t)+3cos⁡(3t))V(t) = e^{-2t}(-2\sin(3t) + 3\cos(3t))

Advanced Chain Rule Examples with Trigonometric Functions

  • Example 1: Differentiating y=(sin⁡(x)−5x3)6cos⁡(x)y = (\sin(x) - 5x^3)^6 \cos(x)

    • Differentiating a composite term (sin⁡(x)−5x3)6(\sin(x) - 5x^3)^6 requires applying the chain rule:     ddx[(sin⁡(x)−5x3)6]=6(sin⁡(x)−5x3)5×(cos⁡(x)−15x2)\frac{d}{dx}\left[(\sin(x) - 5x^3)^6\right] = 6(\sin(x) - 5x^3)^5 \times (\cos(x) - 15x^2)

    • Note on constant distribution: Coefficients outside parenthesis (such as 66) only multiply terms inside their immediate product group and do not distribute across separate additive terms.

  • Example 2: Differentiating double-nested trigonometric functions y=sin⁡(cos⁡(x2))y = \sin(\cos(x^2))

    • Identify the three nested layers: outer function sin⁡(u)\sin(u), middle function cos⁡(v)\cos(v), inner function v=x2v = x^2.

    • Apply the chain rule step-by-step from outermost to innermost:

    • Outer derivative: cos⁡(cos⁡(x2))\cos(\cos(x^2))

    • Middle derivative: −sin⁡(x2)-\sin(x^2)

    • Inner derivative: ddx(x2)=2x\frac{d}{dx}(x^2) = 2x

    • Multiply all parts together:     y′=cos⁡(cos⁡(x2))×(−sin⁡(x2))×2x=−2xsin⁡(x2)cos⁡(cos⁡(x2))y' = \cos(\cos(x^2)) \times (-\sin(x^2)) \times 2x = -2x \sin(x^2) \cos(\cos(x^2))

  • Example 3: Differentiating power of a composite function y=(sin⁡(cos⁡(x)))2y = (\sin(\cos(x)))^2

    • Outer layer is the power rule u2u^2:     ddu(u2)=2u=2sin⁡(cos⁡(x))\frac{d}{du}(u^2) = 2u = 2\sin(\cos(x))

    • Middle layer is derivative of sine:     ddv(sin⁡(v))=cos⁡(v)=cos⁡(cos⁡(x))\frac{d}{dv}(\sin(v)) = \cos(v) = \cos(\cos(x))

    • Inner layer is derivative of cosine:     ddx(cos⁡(x))=−sin⁡(x)\frac{d}{dx}(\cos(x)) = -\sin(x)

    • Full derivative expression:     y′=2sin⁡(cos⁡(x))×cos⁡(cos⁡(x))×(−sin⁡(x))=−2sin⁡(x)sin⁡(cos⁡(x))cos⁡(cos⁡(x))y' = 2\sin(\cos(x)) \times \cos(\cos(x)) \times (-\sin(x)) = -2\sin(x)\sin(\cos(x))\cos(\cos(x))

Graphical and Tabular Chain Rule Problems

  • Example 1: Evaluation of u′(1)u'(1) where u(x)=f(g(x))u(x) = f(g(x)) from a graph

    • Formula:     u′(x)=f′(g(x))×g′(x)  ⟹  u′(1)=f′(g(1))×g′(1)u'(x) = f'(g(x)) \times g'(x) \implies u'(1) = f'(g(1)) \times g'(1)

    • Reading graph values for g(x)g(x) (represented by green line segments):

    • The branch containing x=1x = 1 passes through points (0,6)(0, 6) and (5,−4)(5, -4).

    • Evaluate g(1)g(1): From graph at x=1x = 1, g(1)=4g(1) = 4.

    • Calculate slope g′(1)g'(1) for this linear segment:       g′(1)=m2=y2−y1x2−x1=−4−65−0=−105=−2g'(1) = m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 6}{5 - 0} = \frac{-10}{5} = -2

    • Reading graph values for f(x)f(x) (represented by the red line):

    • Line passes through points (−5,−5)(-5, -5) and (5,5)(5, 5).

    • Calculate slope f′(x)f'(x) across the line:       f′(x)=m1=5−(−5)5−(−5)=1010=1f'(x) = m_1 = \frac{5 - (-5)}{5 - (-5)} = \frac{10}{10} = 1

    • Therefore, f′(4)=1f'(4) = 1.

    • Final calculation for u′(1)u'(1):     u′(1)=f′(4)×g′(1)=1×(−2)=−2u'(1) = f'(4) \times g'(1) = 1 \times (-2) = -2

    • Selection of points note: When computing derivatives on piecewise-linear graphs, points must be chosen exclusively from the specific linear branch containing the evaluation point (x=1x = 1 in this case).

  • Example 2: Evaluation of triple composite function Q′(2)Q'(2) where Q(x)=f(g(h(x)))Q(x) = f(g(h(x)))

    • Apply the chain rule across three layers:     Q′(x)=f′(g(h(x)))×g′(h(x))×h′(x)Q'(x) = f'(g(h(x))) \times g'(h(x)) \times h'(x)

    • Evaluate at x=2x = 2:     Q′(2)=f′(g(h(2)))×g′(h(2))×h′(2)Q'(2) = f'(g(h(2))) \times g'(h(2)) \times h'(2)

    • Given initial values:

    • h(2)=3h(2) = 3

    • h′(2)=5h'(2) = 5

    • g(3)=7g(3) = 7

    • g′(3)=−3g'(3) = -3

    • f′(7)=4f'(7) = 4

    • Step-by-step evaluation:

    • Substitute h(2)=3h(2) = 3 and h′(2)=5h'(2) = 5:       Q′(2)=f′(g(3))×g′(3)×5Q'(2) = f'(g(3)) \times g'(3) \times 5

    • Substitute g(3)=7g(3) = 7 and g′(3)=−3g'(3) = -3:       Q′(2)=f′(7)×(−3)×5Q'(2) = f'(7) \times (-3) \times 5

    • Substitute f′(7)=4f'(7) = 4:       Q′(2)=4×(−3)×5=−60Q'(2) = 4 \times (-3) \times 5 = -60

Differentiation of Exponential Functions

  • Derivation of general exponential rule axa^x:

    • Express axa^x using the natural exponential base ee:     ax=exln⁡(a)a^x = e^{x \ln(a)}

    • Differentiate exln⁡(a)e^{x \ln(a)} with respect to xx using the chain rule:     ddx(ax)=ddx(exln⁡(a))=exln⁡(a)×ln⁡(a)=axln⁡(a)\frac{d}{dx}\left(a^x\right) = \frac{d}{dx}\left(e^{x \ln(a)}\right) = e^{x \ln(a)} \times \ln(a) = a^x \ln(a)

  • Key distinction between Power Rule and Exponential Rule:

    • Power Rule applies when the base is a variable and exponent is a constant (xnx^n):     ddx(xn)=nxn−1\frac{d}{dx}(x^n) = n x^{n-1}

    • Exponential Rule applies when the base is a constant and exponent is a variable (axa^x):     ddx(ax)=axln⁡(a)\frac{d}{dx}(a^x) = a^x \ln(a)

    • Exponents in exponential functions do NOT decrease by 1 upon differentiation.

  • Example 1: Differentiation of y=2x5y = 2^{x^5}

    • Outer function is aua^u where a=2a = 2 and u = x^5$.\n - Apply exponential rule and chain rule:\n    \frac{d}{dx}\left(2^{x^5}\right) = 2^{x^5} \times \ln(2) \times \frac{d}{dx}(x^5)\n - Differentiate inner exponent \frac{d}{dx}(x^5) = 5x^4:\n    \frac{d}{dx}\left(2^{x^5}\right) = 5x^4 \times 2^{x^5} \ln(2)\n\n- Example 2: Differentiation of combined terms x^6 \cdot 6^x + 4^x\n - First term requires Product Rule combined with Exponential Rule:\n    \frac{d}{dx}\left(x^6 \cdot 6^x\right) = \frac{d}{dx}(x^6) \cdot 6^x + x^6 \cdot \frac{d}{dx}(6^x) = 6x^5 \cdot 6^x + x^6 \cdot 6^x \ln(6)\n - Second term requires standard Exponential Rule:\n    \frac{d}{dx}(4^x) = 4^x \ln(4)\n - Total derivative:\n    y' = 6x^5 \cdot 6^x + x^6 \cdot 6^x \ln(6) + 4^x \ln(4)\n\n# Real-World Exponential Growth Application\n\n- Problem Formulation:\n - Population or cost function given by:\n    C(x) = 531353 \times (1.061)^{x - 1990}\n - Target: Find the instantaneous rate of change C'(x)intheyearin the yearx = 2007$.

  • Algebraic Restructuring:

    • Rewrite using exponent rules:     C(x)=(531353×(1.061)−1990)×(1.061)xC(x) = \left(531353 \times (1.061)^{-1990}\right) \times (1.061)^x

  • Differentiation Strategy:

    • Apply exponential differentiation formula ddx(ax−c)=ax−cln⁡(a)\frac{d}{dx}(a^{x - c}) = a^{x - c} \ln(a):     C′(x)=531353×(1.061)x−1990×ln⁡(1.061)C'(x) = 531353 \times (1.061)^{x - 1990} \times \ln(1.061)

    • Evaluate at x=2007x = 2007:     C′(2007)=531353×(1.061)2007−1990×ln⁡(1.061)=531353×(1.061)17×ln⁡(1.061)C'(2007) = 531353 \times (1.061)^{2007 - 1990} \times \ln(1.061) = 531353 \times (1.061)^{17} \times \ln(1.061)

Implicit Differentiation Techniques

  • Definitions and Distinctions:

    • Explicit Function: A function where yy is isolated completely on one side in terms of xx (y=f(x)y = f(x)).

    • Implicit Function: A function where xx and yy are intertwined in an equation such that yy cannot be easily isolated (x2+y2=1x^2 + y^2 = 1).

  • Procedure for Implicit Differentiation:

    • Differentiate both sides of the equation with respect to x$.\n - Treat yasanimplicitfunctionofas an implicit function ofx((y = y(x)).\n - Apply the Chain Rule whenever differentiating terms containing y:\n    \frac{d}{dx}(f(y)) = f'(y) \times y'\n - Group all terms containing y'(or(or\frac{dy}{dx}) on one side of the equation.\n - Factor out y'andsolveforand solve fory' algebraically.\n\n- Execution Example:\n - Differentiating terms of the y-family:\n    \frac{d}{dx}(3y^3) = 9y^2 y'\n    \frac{d}{dx}(-4y) = -4y'\n    \frac{d}{dx}(-y) = -y'\n - Factoring y' out from grouped terms:\n    y'(3y^2 - 4 - 1) = y'(3y^2 - 5)\n\n# Questions & Classroom Discussion\n\n- Question on distributing constant factor 6:\n - Prompt: Should the factor 6 be distributed across the entire derivative expression including subsequent subtracted terms?\n - Answer: No. The constant coefficient 6isattachedonlytothederivativeofitsrespectivetermis attached only to the derivative of its respective term(\sin(x) - 5x^3)^6 up to the subtraction operator. It does not multiply subsequent separate terms.\n\n- Question on point selection for piecewise function derivative from graph:\n - Prompt: When evaluating g'(1),whycanwenotchoosepointslike, why can we not choose points like(-5, -5)andand(-4, -5)onong(x)?\n - Answer: The function g(x)ispiecewiselinearwithdistinctlinesegments(branches).Thepointis piecewise linear with distinct line segments (branches). The pointx = 1liesspecificallyonthebranchconnectinglies specifically on the branch connecting(0, 6)andand(5, -4).Usingpointsfromadifferentbranchyieldstheslopeofthatothersegment,nottheslopeat. Using points from a different branch yields the slope of that other segment, not the slope atx = 1$.

  • Question on differentiation rules for e2te^{2t} vs t2t^2:

    • Prompt: Does differentiating e2te^{2t} drop the power like the power rule?

    • Answer: No. e2te^{2t} is an exponential function, not a power function. The power rule only applies when a variable base is raised to a constant power (t2→2tt^2 \to 2t). Exponential functions retain their exponential structure upon differentiation.