MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 1 of 9 EXAMPLE 1 : Find the derivative of f (x) = 7x β 4 from first principals. Remember: The derivative of the function determines the slope of the tangent. The graph of f (x) = 7x β 4 is a line. The slope of this line is 7 at any point. The derivative is the π(π + π) β π(π) πβπ (π + π) β π π₯π’π¦ f β² (x) = [π(π + π) β π] β (ππ β π) π₯π’π¦ πβπ (π + π) β π f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ π ππ + ππ β π β ππ + π πβπ π ππ πβπ π πβπ f β² (x) = 7 EXAMPLE 2a Find the first derivative ππ¦ ππ₯ of the function y = x2 β 6x β 1 using first principles. [(π + π)π β π(π + π) β π] β (ππ β ππ β π) π π = π₯π’π¦ (π + π) β π π π πβπ π π ππ + πππ + ππ β ππ β ππ β π β ππ + ππ + π = π₯π’π¦ π π πβπ π π π πππ + ππ β ππ = π₯π’π¦ π π πβπ π π π = π₯π’π¦ ππ + π β π π π πβπ π π = ππ β π π π U2 L1 Derivatives ANSWERS f β² (x) = MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 2 of 9 EXAMPLE 2b Find the equation of the line tangent to y = x2 β 6x β 1 at (β 1 ,6) π π = ππ β π π π π π = πππππ ππ πππππππ = π(βπ) β π = βπ π π π = β2 π = βππ β π EXAMPLE 2c π Find the equation of the line tangent to y = x2 β 6x β 1 and perpendicular to π = β π π + π π Slope of perpendicular line to π = β π π + π π’π¬ π π π = ππ β π π π ππ β π = π ππ = π π=π π = ππ β π(π) β π = βπ βπ = π(π) + π π = βππ U2 L1 Derivatives ANSWERS π π=β π+π π π = ππ β ππ MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS EXAMPLE 3: Find f β² (x) if π(π₯) = f β² (x) = Page 3 of 9 7 2 β 3π₯ π π β π β π(π + π) π β ππ π₯π’π¦ πβπ (π + π) β π f β² (x) = f β² (x) = π π β π β π(π + π) π β ππ π₯π’π¦ πβπ (π + π) β π π(π β ππ) β π[π β π(π + π)] πβπ π[π β π(π + π)](π β ππ) π₯π’π¦ ππ β πππ β ππ + πππ + πππ πβπ π[π β π(π + π)](π β ππ) f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ f β² (x) = f β² (x) = πππ πβπ π[π β π(π + π)](π β ππ) ππ πβπ [π β π(π + π)](π β ππ) π₯π’π¦ ππ ππ π₯π’π¦ [π β π(π + π)](π β ππ) f β² (x) = πβπ [π β π(π + π)](π β ππ) f β² (x) = U2 L1 Derivatives ANSWERS ππ (π β ππ)π MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS EXAMPLE 4: Find f β² (x ) of f β² (x) = Page 4 of 9 π(π₯) = βπ₯ + 3 using the limit process βπ + π + π β βπ + π πβπ (π + π) β π π₯π’π¦ βπ + π + π β βπ + π βπ + π + π + βπ + π × πβπ (π + π) β π βπ + π + π + βπ + π f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ f β² (x) = f β² (x) = π + π + π β (π + π) πβπ π(βπ + π + π + βπ + π) π+π+πβπβπ πβπ π(βπ + π + π + βπ + π) π π₯π’π¦ πβπ π(βπ + π + π + βπ + π) π (βπ + π + βπ + π) = π πβ π + π Find the equation of the line tangent to π(π₯) = βπ₯ + 3 π π = f β² (1) = πβ π + π π π= U2 L1 Derivatives ANSWERS π π π·(π, π) π= π (π) + π π π= π π π= π π π+ π π at (1, 2) MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 5 of 9 EXAMPLE 5: Find equation of the line tangent to π(π₯) = βπ₯ β 1 and parallel to π¦ = 2π₯ β 7 f β² (x) = π₯π’π¦ βπ πβπ + π β π β βπ β π (π + π) β π βπ + π β π β βπ β π βπ + π β π + βπ β π × πβπ (π + π) β π βπ + π β π + βπ β π f β² (x) = π₯π’π¦ f β² (x) = π₯π’π¦ π + π β π β (π β π) πβπ π(βπ + f β² (x) = πβ² (π) = π β π + βπ β π) π π₯π’π¦ πβπ π(βπ + π β π + βπ β π) π (βπ + π β π + βπ β π) = π πβπ β π A line parallel to π¦ = ππ₯ + 7 will have a slope of 2 π= π πβπ β π πβπ β π = π π βπ β π = π πβπ= π= π ππ ππ ππ ππ ππ π π π( ) = β βπ = β = ππ ππ ππ π U2 L1 Derivatives ANSWERS π ππ = (π) +π π ππ π= βππ π π = ππ β ππ π MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS QUESTIONS Page 6 of 9 Use first principles to find the derivative of the following function. 1. f(x) = 4x2 β 7 [π(π + π)π β π] β (πππ β π) π π = π₯π’π¦ (π + π) β π π π πβπ π π [π(ππ + πππ + ππ ) β π] β πππ + π = π₯π’π¦ π π πβπ π π π πππ + πππ + πππ β π β πππ + π = π₯π’π¦ π π πβπ π π π πππ + πππ + πππ β π β πππ + π = π₯π’π¦ π π πβπ π π π πππ + πππ = π₯π’π¦ π π πβπ π π π πππ + πππ = π₯π’π¦ π π πβπ π π π = π₯π’π¦ ππ + ππ π π πβπ π π = ππ π π Find the slope of the tangent at (1, β 3) and then write the equation for the tangent. π π = ππ = π(π) = π π π βπ = π(π) + π π = βππ π = ππ β ππ U2 L1 Derivatives ANSWERS MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 7 of 9 2. Find the derivative f β² (x) from first principals π(π₯) = π₯+2 π₯ π+π+π π+π β π πβ² (π) = π₯π’π¦ π + π πβπ (π + π) β π π(π + π + π) β (π + π)(π + π) πβπ ππ(π + π) πβ² (π) = π₯π’π¦ ππ + ππ + ππ β ππ β ππ β ππ β ππ πβπ ππ(π + π) πβ² (π) = π₯π’π¦ βππ πβπ ππ(π + π) πβ² (π) = π₯π’π¦ βπ πβπ π(π + π) πβ² (π) = πβ² (π) = π₯π’π¦ βπ ππ Find the slope of the tangent at the specified point, and then write the equation for the tangent. 3. π(π₯) = π₯+2 π₯ πβ² (π) = at (3, βπ βπ = ππ π π βπ (π) + π = π π π βπ = +π π π π= π π π π π=β π+ π π U2 L1 Derivatives ANSWERS 5 ) 3 MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 8 of 9 4. Find π β² (π₯ ) of π(π₯) = β3π₯ + 1 using the limit process (first principles) βπ(π + π) + π β βππ + π f β² (x) = π₯π’π¦ πβπ (π + π) β π f β² (x) = f β² (x) = f β² (x) = f β² (x) = βππ + ππ + π β βππ + π βππ + ππ + π + βππ + π × πβπ π βππ + ππ + π + βππ + π π₯π’π¦ (ππ + ππ + π) β (ππ + π) π₯π’π¦ πβπ π(βππ + ππ + π + βππ + π ) ππ π₯π’π¦ πβπ π(βππ + π₯π’π¦ ππ + π + βππ + π) π πβπ βππ + ππ + π + βππ + π πβ² (π) = π πβππ + π Find the equation of the line tangent to π(π₯) = β3π₯ + 1 at (5, 4) πβ² (π) = π πβππ + π π π = (π) + π π π= = ππ +π π ππ π π ππ π = π+ π π π= U2 L1 Derivatives ANSWERS π πβπ(π) + π = π π MATH 31 UNIT 2 LESSON #1 DERIVATIVES FROM FIRST PRINCIPLES NAME ANSWERS Page 9 of 9 5 5. Find equation of the line tangent to π(π₯) = β5π₯ β 7 parallel to π¦ = 2 π₯ β 9 βπ(π + π) β π β βππ β π πβπ (π + π) β π f β² (x) = π₯π’π¦ βππ + ππ β π β βππ β π βππ + ππ β π + βππ β π × πβπ π βππ + ππ β π + βππ β π π₯π’π¦ f β² (x) = (ππ + ππ β π) β (ππ β π) π₯π’π¦ f β² (x) = πβπ π(βππ + f β² (x) = ππ π₯π’π¦ πβπ π(βππ + f β² (x) = πβ² (π) = ππ β π + βππ β π) π₯π’π¦ πβπ βππ ππ β π + βππ β π) π + ππ β π + βππ β π π πβππ β π π πβππ β π = π π π π π ( ) = βπ ( ) β π = π π π ππβππ β π = ππ π= βππ β π = π π π ( )+π π π π=π+π ππ β π = π ππ = π π= π π U2 L1 Derivatives ANSWERS π = βπ π π = πβπ π
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