Lab 7: Using Multiple Derivative Rules Study Guide

Laboratory Assignment Overview and Administrative Guidelines

• Assignment Title: LAB 7: USING MULTIPLE DERIVATIVE RULES (ASSIGNMENT)

• Student Identification Requirement: A dedicated section is designated for recording the student's NAME.

• Mandatory Work Requirement:

  • All intermediate mathematical work and reasoning must be fully documented.

  • Correct final answers submitted without accompanying supporting work may receive no credit ($0$ points).

• Algebraic Simplification Policy:

  • Intermediate and final answers do not need to be converted into simplified algebraic forms.

  • Leaving expressions unsimplified is explicitly preferred for evaluation purposes ("in fact, it is better if you do not").

• Total Point Allocation:

  • Problem 1 (Example): $5$ points

  • Problem 2: $10$ points

  • Problem 3: $10$ points

  • Problem 4: $15$ points total (divided into $8$ points for part a, $2$ points for part b, and $5$ points for part c)

  • Combined Total: $40$ points across all four problems

Core Calculus Differentiation Rules Reference

• Power Rule:

  • For any real constant exponent $n$:

  • $\frac{d}{dx}[x^n] = n x^{n-1}$

• Constant Multiple Rule:

  • For any constant $c$:

  • $\frac{d}{dx}[c cdot f(x)] = c cdot f'(x)$

• Sum and Difference Rules:

  • $\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

• Product Rule:

  • For the product of two differentiable functions $f(x)$ and $g(x)$:

  • $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

• Quotient Rule:

  • For the quotient of two differentiable functions with $g(x) \neq 0$:

  • $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

• Chain Rule:

  • For composite functions where $y = f(u)$ and $u = g(x)$:

  • $\frac{d}{dx}[f(g(x))] = f'(g(x)) cdot g'(x)$

• Generalized Power Rule (Power of a Function):

  • For any differentiable inner function $u(x)$:

  • $\frac{d}{dx}[(u(x))^n] = n(u(x))^{n-1} cdot u'(x)$

• Exponential Differentiation Rules:

  • Base $e$ (Natural Exponential Function):

  • $\frac{d}{dx}[e^{u(x)}] = e^{u(x)} cdot u'(x)$

  • Base $a$ ($a > 0$, $a \neq 1$):

  • $\frac{d}{dx}[a^{u(x)}] = a^{u(x)} cdot \ln(a) cdot u'(x)$

• Logarithmic Differentiation Rules:

  • Base $e$ (Natural Logarithmic Function):

  • $\frac{d}{dx}[\ln(u(x))] = \frac{1}{u(x)} cdot u'(x)$

  • Base $a$ ($a > 0$, $a \neq 1$):

• $\frac{d}{dx}[\log_a(u(x))] = \frac{1}{u(x) cdot \ln(a)} cdot u'(x)$

Problem 1: Composite Square Root and Quotient Rule (Example - 5 Points)

• Problem Statement:

  • If $h(x) = \sqrt{\frac{e^{3x}}{x^2 + 1}}$, then compute $h'(x)$.

• Structural Analysis:

  • Outer Function: Radical expression rewritten as an exponent of $\frac{1}{2}$:

  • $h(x) = \left(\frac{e^{3x}}{x^2 + 1}\right)^{1/2}$

  • Inner Function: Rational expression consisting of an exponential numerator and a polynomial denominator:

  • $u(x) = \frac{e^{3x}}{x^2 + 1}$

• Step-by-Step Differentiation Procedure:

  • Step 1: Apply the Generalized Power Rule to the outer shell:

  • $\frac{d}{du}[u^{1/2}] = \frac{1}{2} u^{-1/2} = \frac{1}{2} \left(\frac{e^{3x}}{x^2 + 1}\right)^{-1/2}$

  • Step 2: Differentiate the inner quotient $u(x) = \frac{N(x)}{D(x)}$ using the Quotient Rule:

  • Numerator function: $N(x) = e^{3x}$

  • Derivative of numerator by Exponential Chain Rule: $N'(x) = 3e^{3x}$

  • Denominator function: $D(x) = x^2 + 1$

  • Derivative of denominator by Power Rule: $D'(x) = 2x$

  • Quotient derivative expression:

  • $u'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2} = \frac{3e^{3x}(x^2 + 1) - e^{3x}(2x)}{(x^2 + 1)^2}$

  • Step 3: Combine outer and inner derivatives via the Chain Rule:

  • $h'(x) = \frac{1}{2}\left(\frac{e^{3x}}{x^2 + 1}\right)^{-1/2} cdot \left[\frac{3e^{3x}(x^2 + 1) - e^{3x}(2x)}{(x^2 + 1)^2}\right]$

• Final Answer in Preferred Unsimplified Form:

  • $h'(x) = \frac{1}{2}\left(\frac{e^{3x}}{x^2 + 1}\right)^{-1/2} \left(\frac{3e^{3x}(x^2 + 1) - 2x e^{3x}}{(x^2 + 1)^2}\right)$

Problem 2: Logarithmic and Radical Quotient Rule (10 Points)

• Problem Statement:

  • If $k(x) = \frac{\log_7(x)}{\sqrt[3]{7x - 1}}$, then compute $k'(x)$.

• Structural Analysis:

  • Overarching form: A quotient $k(x) = \frac{f(x)}{g(x)}$.

  • Numerator: Base-$7$ logarithm function:

  • $f(x) = \log_7(x)$

  • Denominator: Cube root rewritten in rational exponent form:

  • $g(x) = (7x - 1)^{1/3}$

• Step-by-Step Differentiation Procedure:

  • Step 1: Differentiate the numerator $f(x)$ using the Base-$a$ Logarithm Rule:

  • $f'(x) = \frac{1}{x \ln(7)}$

  • Step 2: Differentiate the denominator $g(x)$ using the Chain Rule:

  • Outer derivative with respect to inner quantity: $\frac{1}{3}(7x - 1)^{-2/3}$

  • Inner derivative: $\frac{d}{dx}[7x - 1] = 7$

  • Resulting derivative:

  • $g'(x) = \frac{1}{3}(7x - 1)^{-2/3} cdot 7 = \frac{7}{3}(7x - 1)^{-2/3}$

  • Step 3: Compute the denominator squared:

  • $[g(x)]^2 = [(7x - 1)^{1/3}]^2 = (7x - 1)^{2/3}$

  • Step 4: Assemble using the Quotient Rule formula:

  • $k'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

  • $k'(x) = \frac{\left(\frac{1}{x \ln(7)}\right)(7x - 1)^{1/3} - \log_7(x)\left(\frac{7}{3}(7x - 1)^{-2/3}\right)}{(7x - 1)^{2/3}}$

• Alternative Expression Variation (Denominator with Exponential Term $7^x - 1$):

  • If the denominator represents $\sqrt[3]{7^x - 1} = (7^x - 1)^{1/3}$:

  • Inner derivative: $\frac{d}{dx}[7^x - 1] = 7^x \ln(7)$

  • Denominator derivative: $g'(x) = \frac{1}{3}(7^x - 1)^{-2/3} cdot 7^x \ln(7)$

  • Derivative expression:

  • $k'(x) = \frac{\left(\frac{1}{x \ln(7)}\right)(7^x - 1)^{1/3} - \log_7(x)\left(\frac{1}{3}(7^x - 1)^{-2/3} 7^x \ln(7)\right)}{(7^x - 1)^{2/3}}$

• Alternative Expression Variation (Denominator as $\sqrt{3} cdot 7^x - 1$):

  • If the denominator represents $\sqrt{3} cdot 7^x - 1$:

  • Denominator derivative: $g'(x) = \sqrt{3} cdot 7^x \ln(7)$

  • Derivative expression:

  • $k'(x) = \frac{\left(\frac{1}{x \ln(7)}\right)(\sqrt{3} cdot 7^x - 1) - \log_7(x)(\sqrt{3} cdot 7^x \ln(7))}{(\sqrt{3} cdot 7^x - 1)^2}$

• Final Answer in Preferred Unsimplified Form (Standard Reading):

  • $k'(x) = \frac{\frac{1}{x \ln(7)}(7x - 1)^{1/3} - \log_7(x) \left[\frac{7}{3}(7x - 1)^{-2/3}\right]}{(7x - 1)^{2/3}}$

Problem 3: Fifth-Root Composite Product Rule (10 Points)

• Problem Statement:

  • If $\ell(x) = \sqrt[5]{e^{4x} \ln(x^2 - 2x)}$, then compute $\ell'(x)$.

• Structural Analysis:

  • Outer Structure: A fifth root, expressed as a power of $\frac{1}{5}$:

  • $\ell(x) = [e^{4x} \ln(x^2 - 2x)]^{1/5}$

  • Inner Structure: A product of two variable factors, $u(x) = f(x) cdot g(x)$, where:

  • First factor: $f(x) = e^{4x}$

  • Second factor: $g(x) = \ln(x^2 - 2x)$

• Step-by-Step Differentiation Procedure:

  • Step 1: Differentiate the outer power function:

  • $\frac{d}{du}[u^{1/5}] = \frac{1}{5} u^{-4/5} = \frac{1}{5} [e^{4x} \ln(x^2 - 2x)]^{-4/5}$

  • Step 2: Differentiate the inner product $u(x) = f(x)g(x)$ using the Product Rule:

  • Derivative of first factor $f(x) = e^{4x}$ via Chain Rule: $f'(x) = 4e^{4x}$

  • Derivative of second factor $g(x) = \ln(x^2 - 2x)$ via Logarithmic Chain Rule:

  • $g'(x) = \frac{1}{x^2 - 2x} cdot \frac{d}{dx}[x^2 - 2x] = \frac{2x - 2}{x^2 - 2x}$

  • Product Rule assembly for the inner derivative:

  • $u'(x) = f'(x)g(x) + f(x)g'(x) = 4e^{4x} \ln(x^2 - 2x) + e^{4x}\left(\frac{2x - 2}{x^2 - 2x}\right)$

  • Step 3: Combine outer and inner derivatives via the Chain Rule:

  • $\ell'(x) = \frac{1}{5} [e^{4x} \ln(x^2 - 2x)]^{-4/5} cdot \left[4e^{4x} \ln(x^2 - 2x) + e^{4x}\left(\frac{2x - 2}{x^2 - 2x}\right)\right]$

• Final Answer in Preferred Unsimplified Form:

  • $\ell'(x) = \frac{1}{5}[e^{4x}\ln(x^2 - 2x)]^{-4/5} \left(4e^{4x}\ln(x^2 - 2x) + e^{4x} \left[\frac{2x - 2}{x^2 - 2x}\right]\right)$

Problem 4: Rational Derivative, Evaluation, and Tangent Line (15 Points Total)

• Function Definition:

  • Consider $m(x) = \frac{\log_3(x^4 - x^2 + 3)}{\sqrt[3]{3x^6 - 6x + 11}}$

  • Rewritten in power notation:

  • $m(x) = \frac{\log_3(x^4 - x^2 + 3)}{(3x^6 - 6x + 11)^{1/3}}$

• Part a: Compute $m'(x)$ (8 Points):

  • Overarching Differentiation Rule: Quotient Rule:

  • $m'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

  • Differentiation of Numerator $u(x) = \log_3(x^4 - x^2 + 3)$:

  • Base-$a$ logarithm rule combined with polynomial chain rule:

  • $u'(x) = \frac{1}{(x^4 - x^2 + 3)\ln(3)} cdot \frac{d}{dx}[x^4 - x^2 + 3] = \frac{4x^3 - 2x}{(x^4 - x^2 + 3)\ln(3)}$

  • Differentiation of Denominator $v(x) = (3x^6 - 6x + 11)^{1/3}$:

  • Generalized power rule combined with polynomial chain rule:

  • $v'(x) = \frac{1}{3}(3x^6 - 6x + 11)^{-2/3} cdot \frac{d}{dx}[3x^6 - 6x + 11] = \frac{1}{3}(3x^6 - 6x + 11)^{-2/3}(18x^5 - 6)$

  • Denominator squared:

  • $[v(x)]^2 = [(3x^6 - 6x + 11)^{1/3}]^2 = (3x^6 - 6x + 11)^{2/3}$

  • Synthesis into the complete unsimplified derivative:

  • $m'(x) = \frac{\left[\frac{4x^3 - 2x}{(x^4 - x^2 + 3)\ln(3)}\right](3x^6 - 6x + 11)^{1/3} - \log_3(x^4 - x^2 + 3)\left[\frac{1}{3}(3x^6 - 6x + 11)^{-2/3}(18x^5 - 6)\right]}{(3x^6 - 6x + 11)^{2/3}}$

• Part b: Evaluate $m(1)$ and $m'(1)$ (2 Points):

  • Evaluation of $m(1)$:

  • Numerator evaluation at $x = 1$:

  • $u(1) = \log_3(1^4 - 1^2 + 3) = \log_3(1 - 1 + 3) = \log_3(3) = 1$

  • Denominator evaluation at $x = 1$:

  • $v(1) = \sqrt[3]{3(1)^6 - 6(1) + 11} = \sqrt[3]{3 - 6 + 11} = \sqrt[3]{8} = 2$

  • Value of the original function at $x = 1$:

  • $m(1) = \frac{u(1)}{v(1)} = \frac{1}{2}$

  • Evaluation of $m'(1)$:

  • Evaluate $u'(1)$:

  • $u'(1) = \frac{4(1)^3 - 2(1)}{(1^4 - 1^2 + 3)\ln(3)} = \frac{4 - 2}{(3)\ln(3)} = \frac{2}{3\ln(3)}$

  • Evaluate $v'(1)$:

  • $v'(1) = \frac{1}{3}(8)^{-2/3}(18(1)^5 - 6) = \frac{1}{3} \left(\frac{1}{8^{2/3}}\right)(12) = \frac{1}{3} \left(\frac{1}{4}\right)(12) = \frac{12}{12} = 1$

  • Evaluate $[v(1)]^2$:

  • $[v(1)]^2 = 2^2 = 4$

  • Substitute evaluated components into the Quotient Rule formula:

  • $m'(1) = \frac{u'(1)v(1) - u(1)v'(1)}{[v(1)]^2} = \frac{\left(\frac{2}{3\ln(3)}\right)(2) - (1)(1)}{4}$

  • Numerator algebraic simplification:

  • $\frac{4}{3\ln(3)} - 1 = \frac{4 - 3\ln(3)}{3\ln(3)}$

  • Final value for $m'(1)$:

  • $m'(1) = \frac{\frac{4 - 3\ln(3)}{3\ln(3)}}{4} = \frac{4 - 3\ln(3)}{12\ln(3)}$

  • Alternative split fraction form:

  • $m'(1) = \frac{1}{3\ln(3)} - \frac{1}{4}$

• Part c: Find the Equation of the Tangent Line (5 Points):

  • Point of tangency coordinates:

  • $(x_0, y_0) = (1, m(1)) = \left(1, \frac{1}{2}\right)$

  • Slope of the tangent line ($M$):

  • $M = m'(1) = \frac{4 - 3\ln(3)}{12\ln(3)}$

  • Standard Point-Slope Form equation:

  • $y - y_0 = M(x - x_0)$

  • $y - \frac{1}{2} = \left(\frac{4 - 3\ln(3)}{12\ln(3)}\right)(x - 1)$

  • Conversion to Slope-Intercept Form ($y = Mx + b$):

  • Distribute the slope $M$ across $(x - 1)$:

  • $y = \left(\frac{4 - 3\ln(3)}{12\ln(3)}\right)x - \left(\frac{4 - 3\ln(3)}{12\ln(3)}\right) + \frac{1}{2}$

  • Combine constant terms with common denominator $12\ln(3)$:

  • $b = -\left(\frac{4 - 3\ln(3)}{12\ln(3)}\right) + \frac{6\ln(3)}{12\ln(3)} = \frac{-4 + 3\ln(3) + 6\ln(3)}{12\ln(3)} = \frac{9\ln(3) - 4}{12\ln(3)}$

  • Resulting Slope-Intercept equation:

  • $y = \left(\frac{4 - 3\ln(3)}{12\ln(3)}\right)x + \frac{9\ln(3) - 4}{12\ln(3)}$