Examkin

Differentiation from First Principles

22 questions

All types22Multiple choice11Numeric7Short answer4
Question 21Question →

By definition, the derivative of the function f(x)=4xf(x) = \frac{4}{x} at x=2x = 2 is given by the limit of the difference quotient lim⁡h→0f(2+h)−f(2)h\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}. What is the value of this limit?

Show answer & explanation

Answer: −1-1

Answer

−1-1
Substituting f(2+h)=42+hf(2+h) = \frac{4}{2+h} and f(2)=2f(2) = 2 into the difference quotient gives 42+h−2h=−2hh(2+h)=−22+h\frac{\frac{4}{2+h} - 2}{h} = \frac{-2h}{h(2+h)} = \frac{-2}{2+h}. Evaluating the limit as h→0h \to 0 yields −22=−1\frac{-2}{2} = -1.

Step-by-Step Solution

1
Calculate f(2)f(2) and f(2+h)f(2+h)
f(2)=42=2f(2) = \frac{4}{2} = 2 and f(2+h)=42+hf(2+h) = \frac{4}{2+h}
These are the two values needed for the difference quotient numerator.
2
Subtract f(2)f(2) from f(2+h)f(2+h) and find a common denominator
f(2+h)−f(2)=42+h−2=4−2(2+h)2+h=4−4−2h2+h=−2h2+hf(2+h) - f(2) = \frac{4}{2+h} - 2 = \frac{4 - 2(2+h)}{2+h} = \frac{4 - 4 - 2h}{2+h} = \frac{-2h}{2+h}
Simplifying the numerator expression algebraically.
3
Divide the numerator by hh to form the difference quotient
−2h2+hh=−22+h\frac{\frac{-2h}{2+h}}{h} = \frac{-2}{2+h}
Canceling the common factor hh in the numerator and denominator.
4
Evaluate the limit as h→0h \to 0
\lim_{h \to 0} \frac{-2}{2+h} = \frac{-2}{2+0} = -1
Direct substitution of h=0h = 0 into the simplified expression gives the instantaneous rate of change.

Key Concept

Differentiation of reciprocal functions from first principles
Estimated Time:1m 30s
Question 22Question →

Given the function f(x)=5x2−4x+3f(x) = 5x^2 - 4x + 3, what is the numerical value of its derivative at x=2x = 2 when evaluated using the first principles limit definition lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}?

Show answer & explanation

Answer: 16

Answer

The numerical value of the derivative of f(x)=5x2−4x+3f(x) = 5x^2 - 4x + 3 at x=2x = 2 is 16.
Evaluating the first principles definition lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} for f(x)=5x2−4x+3f(x) = 5x^2 - 4x + 3 yields the derivative f′(x)=10x−4f'(x) = 10x - 4. Substituting x=2x = 2 yields 10(2)−4=1610(2) - 4 = 16.

Step-by-Step Solution

1
Determine the expanded form of f(x+h)f(x+h)
f(x+h)=5(x+h)2−4(x+h)+3=5x2+10xh+5h2−4x−4h+3f(x+h) = 5(x+h)^2 - 4(x+h) + 3 = 5x^2 + 10xh + 5h^2 - 4x - 4h + 3
Evaluating the function at x+hx+h requires expanding the square and distributing the constant factors.
2
Compute the difference f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=(5x2+10xh+5h2−4x−4h+3)−(5x2−4x+3)=10xh+5h2−4hf(x+h) - f(x) = (5x^2 + 10xh + 5h^2 - 4x - 4h + 3) - (5x^2 - 4x + 3) = 10xh + 5h^2 - 4h
Subtracting f(x)f(x) cancels terms independent of hh.
3
Form and simplify the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h}
10xh+5h2−4hh=10x+5h−4\frac{10xh + 5h^2 - 4h}{h} = 10x + 5h - 4
Factoring hh out of the numerator allows division by hh for non-zero hh.
4
Evaluate the limit as h→0h \to 0
f′(x)=lim⁡h→0(10x+5h−4)=10x−4f'(x) = \lim_{h \to 0} (10x + 5h - 4) = 10x - 4
Taking the limit produces the general derivative function dydx\frac{\mathrm{d}y}{\mathrm{d}x}.
5
Substitute x=2x = 2 into the derivative
f′(2)=10(2)−4=16f'(2) = 10(2) - 4 = 16
Evaluating at x=2x = 2 gives the instantaneous rate of change at that point.

Key Concept

Differentiation from First Principles
Estimated Time:1m 30s
PreviousPage 2 / 2
Differentiation from First Principles Practice Questions — JAMB UTME — Page 2 | Examkin