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Differentiation from First Principles

22 questions

All types22Multiple choice11Numeric7Short answer4
Question 1Question →

Using the first principles of differentiation for the reciprocal function f(x)=4xf(x) = \frac{4}{x}, evaluate the limit of the difference quotient as h→0h \to 0: lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. What is the simplified expression for the derivative dydx\frac{dy}{dx}?

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Answer: -\frac{4}{x^2}; -4/x^2; -4 / x^2; -\frac{4}{x^{2}}

Answer

The derivative dydx\frac{dy}{dx} is −4x2-\frac{4}{x^2}.
Substituting f(x)=4xf(x) = \frac{4}{x} into the first principles limit formula yields lim⁡h→04x−4(x+h)h⋅x(x+h)=lim⁡h→0−4hh⋅x(x+h)\lim_{h \to 0} \frac{4x - 4(x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-4h}{h \cdot x(x+h)}. Canceling hh gives lim⁡h→0−4x(x+h)=−4x2\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x^2}.

Step-by-Step Solution

1
Write the first principles formula and substitute f(x)=4xf(x) = \frac{4}{x}.
\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}
Apply the definition of differentiation from first principles.
2
Combine the fractions in the numerator using a common denominator x(x+h)x(x+h).
\frac{4}{x+h} - \frac{4}{x} = \frac{4x - 4(x+h)}{x(x+h)} = \frac{4x - 4x - 4h}{x(x+h)} = \frac{-4h}{x(x+h)}
Simplify the numerator into a single fractional expression.
3
Divide the simplified numerator by hh and cancel the common factor of hh.
\frac{\frac{-4h}{x(x+h)}}{h} = \frac{-4h}{h \cdot x(x+h)} = -\frac{4}{x(x+h)}
Eliminate the indeterminate factor of hh from the denominator.
4
Evaluate the limit as h→0h \to 0.
\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x(x+0)} = -\frac{4}{x^2}
Substitute h=0h = 0 into the simplified algebraic expression.

Key Concept

Differentiation from First Principles
Question 2Question →

In differentiating the quadratic function f(x)=2x2+5xf(x) = 2x^2 + 5x from first principles, which of the following expressions represents the fully simplified difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} prior to evaluating the limit as h→0h \to 0?

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Answer: 4x+2h+54x + 2h + 5

Answer

The fully simplified difference quotient is 4x+2h+54x + 2h + 5.
Expanding f(x+h)=2(x+h)2+5(x+h)f(x+h) = 2(x+h)^2 + 5(x+h) gives 2x2+4xh+2h2+5x+5h2x^2 + 4xh + 2h^2 + 5x + 5h. Subtracting f(x)=2x2+5xf(x) = 2x^2 + 5x leaves 4xh+2h2+5h4xh + 2h^2 + 5h. Dividing this expression term-by-term by hh yields 4x+2h+54x + 2h + 5.

Step-by-Step Solution

1
Evaluate f(x+h)f(x+h) for the function f(x)=2x2+5xf(x) = 2x^2 + 5x.
f(x+h)=2(x+h)2+5(x+h)=2(x2+2xh+h2)+5x+5h=2x2+4xh+2h2+5x+5hf(x+h) = 2(x+h)^2 + 5(x+h) = 2(x^2 + 2xh + h^2) + 5x + 5h = 2x^2 + 4xh + 2h^2 + 5x + 5h
To apply first principles, we must replace every instance of xx with (x+h)(x+h) and expand completely.
2
Form the numerator of the difference quotient, f(x+h)−f(x)f(x+h) - f(x).
f(x+h)−f(x)=(2x2+4xh+2h2+5x+5h)−(2x2+5x)=4xh+2h2+5hf(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 5x + 5h) - (2x^2 + 5x) = 4xh + 2h^2 + 5h
Subtracting f(x)f(x) cancels out the terms 2x22x^2 and 5x5x that do not contain hh.
3
Divide the result by hh to get the difference quotient.
\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5
Factoring out hh from the numerator and canceling it with the denominator simplifies the quotient.

Key Concept

Differentiation from First Principles (Difference Quotient)
Estimated Time:1m 30s
Question 3Question →

Using differentiation from first principles, what is the numerical value of the derivative of the function f(x)=3x2−4x+1f(x) = 3x^2 - 4x + 1 at x=2x = 2?

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Answer: 8

Answer

The numerical value of the derivative of f(x)=3x2−4x+1f(x) = 3x^2 - 4x + 1 at x=2x = 2 is 8.
Using the first-principles limit definition, the increment f(2+h)−f(2)f(2+h) - f(2) simplifies to 8h+3h28h + 3h^2. Dividing by hh yields 8+3h8 + 3h, which evaluates to 8 as h→0h \to 0.

Step-by-Step Solution

1
Calculate f(2)f(2)
f(2)=5f(2) = 5
Substitute x=2x = 2 into f(x)=3x2−4x+1f(x) = 3x^2 - 4x + 1.
2
Expand f(2+h)f(2+h)
f(2+h)=5+8h+3h2f(2+h) = 5 + 8h + 3h^2
Substitute x=2+hx = 2+h into f(x)f(x) and expand algebraically.
3
Simplify the difference quotient f(2+h)−f(2)h\frac{f(2+h) - f(2)}{h}
8h+3h2h=8+3h\frac{8h + 3h^2}{h} = 8 + 3h
Subtract f(2)f(2) from f(2+h)f(2+h) and divide every term by hh.
4
Evaluate the limit as h→0h \to 0
f′(2)=8f'(2) = 8
As hh approaches 0, the term 3h3h vanishes, leaving 8.

Key Concept

Differentiation from First Principles
Estimated Time:1m 30s
Question 4Question →

Using differentiation from first principles, what is the derivative of the function f(x)=x2+3xf(x) = x^2 + 3x with respect to xx?

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Answer: 2x+32x + 3

Answer

2x+32x + 3
Applying the first principles limit formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} gives x2+2xh+h2+3x+3h−(x2+3x)h=2xh+h2+3hh=2x+h+3\frac{x^2 + 2xh + h^2 + 3x + 3h - (x^2 + 3x)}{h} = \frac{2xh + h^2 + 3h}{h} = 2x + h + 3. As h→0h \to 0, this expression evaluates to 2x+32x + 3.

Step-by-Step Solution

1
Set up the difference quotient using the definition of differentiation from first principles
\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 + 3(x+h)] - [x^2 + 3x]}{h}
The definition of derivative from first principles requires finding lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
2
Expand algebraic terms in the numerator
\frac{x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x}{h}
Expanding (x+h)2=x2+2xh+h2(x+h)^2 = x^2 + 2xh + h^2 and 3(x+h)=3x+3h3(x+h) = 3x + 3h allows terms to be combined.
3
Simplify the numerator by canceling like terms and dividing by hh
\frac{2xh + h^2 + 3h}{h} = 2x + h + 3
The terms x2−x2=0x^2 - x^2 = 0 and 3x−3x=03x - 3x = 0 cancel out, leaving terms containing hh, which can be divided by hh.
4
Evaluate the limit as h→0h \to 0
\lim_{h \to 0} (2x + h + 3) = 2x + 3
Setting h=0h = 0 in the simplified quotient yields the final derivative f′(x)=2x+3f'(x) = 2x + 3.

Key Concept

Differentiation from first principles
Question 5Question →

Using differentiation from first principles, evaluate the value of the derivative dydx=lim⁡h→0f(x+h)−f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} for the cubic function f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 at the point x=3x = 3.

Show answer & explanation

Answer: 12

Answer

The derivative evaluated at x=3x = 3 is equal to 12.
Evaluating the definition of the derivative from first principles for f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 yields lim⁡h→0f(x+h)−f(x)h=6x2−18x+12\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 6x^2 - 18x + 12. Substituting x=3x = 3 gives 6(3)2−18(3)+12=54−54+12=126(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12.

Step-by-Step Solution

1
Set up the difference quotient definition from first principles
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Differentiation from first principles requires finding the limit of the average rate of change as the increment hh approaches zero.
2
Substitute (x+h)(x+h) into f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5 and expand
f(x+h)=2x3+6x2h+6xh2+2h3−9x2−18xh−9h2+12x+12h−5f(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 9x^2 - 18xh - 9h^2 + 12x + 12h - 5
Expanding binomial terms (x+h)3(x+h)^3 and (x+h)2(x+h)^2 reveals all components involving hh.
3
Calculate f(x+h)−f(x)f(x+h) - f(x) and factor out hh
f(x+h)−f(x)=h(6x2+6xh+2h2−18x−9h+12)f(x+h) - f(x) = h(6x^2 + 6xh + 2h^2 - 18x - 9h + 12)
Terms independent of hh cancel out completely, isolating hh as a common factor.
4
Divide by hh and evaluate the limit as h→0h \to 0
f′(x)=6x2−18x+12f'(x) = 6x^2 - 18x + 12
Canceling hh resolves the 00\frac{0}{0} indeterminate form, allowing direct substitution of h=0h=0.
5
Substitute x=3x = 3 into f′(x)f'(x)
f′(3)=6(3)2−18(3)+12=12f'(3) = 6(3)^2 - 18(3) + 12 = 12
Evaluating at x=3x = 3 gives the numerical value of the instantaneous rate of change at that specific point.

Key Concept

Differentiation from First Principles
Question 6Question →

Using differentiation from first principles, what is the derivative of the function f(x)=3xf(x) = \frac{3}{x} for x≠0x \neq 0?

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Answer: −3x2-\frac{3}{x^2}

Answer

−3x2-\frac{3}{x^2}
Using the first principles formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, substituting f(x)=3xf(x) = \frac{3}{x} gives lim⁡h→03x+h−3xh=lim⁡h→0−3hhx(x+h)=−3x2\lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h} = \lim_{h \to 0} \frac{-3h}{h x(x+h)} = -\frac{3}{x^2}.

Step-by-Step Solution

1
Set up the first principles difference quotient
f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→03x+h−3xhf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h}
Definition of differentiation from first principles.
2
Combine the fractions in the numerator using a common denominator
3x+h−3x=3x−3(x+h)x(x+h)=3x−3x−3hx(x+h)=−3hx(x+h)\frac{3}{x+h} - \frac{3}{x} = \frac{3x - 3(x+h)}{x(x+h)} = \frac{3x - 3x - 3h}{x(x+h)} = \frac{-3h}{x(x+h)}
Algebraic fraction subtraction.
3
Divide by hh and cancel the common factor
−3hx(x+h)h=−3x(x+h)\frac{\frac{-3h}{x(x+h)}}{h} = \frac{-3}{x(x+h)}
Simplifying the difference quotient before taking the limit.
4
Evaluate the limit as h→0h \to 0
\lim_{h \to 0} \frac{-3}{x(x+h)} = \frac{-3}{x(x+0)} = -\frac{3}{x^2}
Direct substitution of h=0h=0 after canceling hh.

Key Concept

Differentiation of a reciprocal function using the first principles limit definition.
Question 7Question →

Using differentiation from first principles, what is the derivative dydx\frac{\mathrm{d}y}{\mathrm{d}x} of the function f(x)=3x2−2xf(x) = 3x^2 - 2x?

Show answer & explanation

Answer: 6x−26x - 2

Answer

The derivative of f(x)=3x2−2xf(x) = 3x^2 - 2x with respect to xx is 6x−26x - 2.
Applying the first principles definition lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to f(x)=3x2−2xf(x) = 3x^2 - 2x yields 3x2+6xh+3h2−2x−2h−(3x2−2x)h=6xh+3h2−2hh=6x+3h−2\frac{3x^2 + 6xh + 3h^2 - 2x - 2h - (3x^2 - 2x)}{h} = \frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2. As h→0h \to 0, this simplifies directly to 6x−26x - 2.

Step-by-Step Solution

1
Set up the definition of the derivative from first principles.
dydx=lim⁡h→0f(x+h)−f(x)h\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
The definition of the derivative is the limit of the difference quotient as hh approaches zero.
2
Substitute f(x+h)f(x+h) and f(x)f(x) into the difference quotient.
f(x+h)=3(x+h)2−2(x+h)=3(x2+2xh+h2)−2x−2h=3x2+6xh+3h2−2x−2hf(x+h) = 3(x+h)^2 - 2(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h = 3x^2 + 6xh + 3h^2 - 2x - 2h
Expand (x+h)2(x+h)^2 completely using algebraic identity.
3
Subtract f(x)f(x) from f(x+h)f(x+h) to find the numerator.
f(x+h)−f(x)=(3x2+6xh+3h2−2x−2h)−(3x2−2x)=6xh+3h2−2hf(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) = 6xh + 3h^2 - 2h
Cancel out identical terms 3x23x^2 and −2x-2x.
4
Divide by hh and evaluate the limit as h→0h \to 0.
\frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2; \quad \lim_{h \to 0} (6x + 3h - 2) = 6x - 2
Factor out hh to cancel the denominator, then set h=0h = 0.

Key Concept

Differentiation from First Principles
Estimated Time:1m 30s
Question 8Question →

When evaluating the derivative of the cubic function f(x)=2x3−5xf(x) = 2x^3 - 5x from first principles, which of the following expressions represents the fully simplified difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} prior to taking the limit as h→0h \to 0?

Show answer & explanation

Answer: 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5

Answer

The simplified difference quotient before evaluating the limit is 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5.
Expanding f(x+h)=2(x+h)3−5(x+h)f(x+h) = 2(x+h)^3 - 5(x+h) using the binomial expansion yields 2x3+6x2h+6xh2+2h3−5x−5h2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h. Subtracting f(x)=2x3−5xf(x) = 2x^3 - 5x leaves 6x2h+6xh2+2h3−5h6x^2h + 6xh^2 + 2h^3 - 5h. Dividing each term by hh gives the correct simplified difference quotient 6x2+6xh+2h2−56x^2 + 6xh + 2h^2 - 5.

Step-by-Step Solution

1
Express f(x+h)f(x+h) for the function f(x)=2x3−5xf(x) = 2x^3 - 5x
f(x+h)=2(x+h)3−5(x+h)=2(x3+3x2h+3xh2+h3)−5x−5h=2x3+6x2h+6xh2+2h3−5x−5hf(x+h) = 2(x+h)^3 - 5(x+h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5x - 5h = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h
Substitute (x+h)(x+h) into f(x)f(x) and expand the binomial terms completely.
2
Subtract f(x)f(x) from f(x+h)f(x+h) to find the numerator f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=(2x3+6x2h+6xh2+2h3−5x−5h)−(2x3−5x)=6x2h+6xh2+2h3−5hf(x+h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h
Cancel out the common terms 2x32x^3 and −5x-5x.
3
Divide the numerator by hh to form the difference quotient
\frac{6x^2h + 6xh^2 + 2h^3 - 5h}{h} = 6x^2 + 6xh + 2h^2 - 5
Factor out hh from each term in the numerator and divide by hh.

Key Concept

Differentiation from First Principles
Question 9Question →

Using differentiation from first principles, evaluate the numerical value of the derivative of the polynomial function f(x)=2x3−3x2+4f(x) = 2x^3 - 3x^2 + 4 at x=2x = 2.

Show answer & explanation

Answer: 12

Answer

The numerical value of the derivative at x=2x = 2 is 12.
Applying the first principles limit formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to f(x)=2x3−3x2+4f(x) = 2x^3 - 3x^2 + 4 yields f′(x)=6x2−6xf'(x) = 6x^2 - 6x. Evaluating this derivative function at x=2x = 2 yields 6(4)−12=126(4) - 12 = 12.

Step-by-Step Solution

1
Set up the difference quotient using the definition of differentiation from first principles
f′(x)=lim⁡h→0[2(x+h)3−3(x+h)2+4]−[2x3−3x2+4]hf'(x) = \lim_{h \to 0} \frac{[2(x+h)^3 - 3(x+h)^2 + 4] - [2x^3 - 3x^2 + 4]}{h}
Differentiation from first principles evaluates the limit of the rate of change as the increment hh approaches zero.
2
Expand (x+h)3(x+h)^3 and (x+h)2(x+h)^2 and subtract f(x)f(x)
f(x+h)−f(x)=6x2h+6xh2+2h3−6xh−3h2f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 6xh - 3h^2
Expanding the terms allows cancellation of all terms not containing hh.
3
Divide by hh and evaluate the limit as h→0h \to 0
f′(x)=lim⁡h→0(6x2+6xh+2h2−6x−3h)=6x2−6xf'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 6x - 3h) = 6x^2 - 6x
Dividing by hh eliminates the indeterminate form 00\frac{0}{0}, allowing direct substitution of h=0h = 0.
4
Substitute x=2x = 2 into the derivative function f′(x)f'(x)
f′(2)=6(2)2−6(2)=24−12=12f'(2) = 6(2)^2 - 6(2) = 24 - 12 = 12
Evaluating at the given point gives the slope of the tangent line at x=2x = 2.

Key Concept

Differentiation from first principles using limit of difference quotient
Question 10Question →

Using differentiation from first principles, evaluate the value of the derivative of the function f(x)=2x2+3x−1f(x) = 2x^2 + 3x - 1 at the point where x=1x = 1.

Show answer & explanation

Answer: 7

Answer

The value of the derivative at x=1x = 1 is 7.
Using first principles, the derivative is defined as f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. Evaluating at x=1x = 1, we find f(1)=4f(1) = 4 and f(1+h)=4+7h+2h2f(1+h) = 4 + 7h + 2h^2. Subtracting f(1)f(1) leaves 7h+2h27h + 2h^2, and dividing by hh yields 7+2h7 + 2h. Taking the limit as h→0h \to 0 yields the final result of 7.

Step-by-Step Solution

1
Calculate the value of the function at x=1x = 1
f(1)=2(1)2+3(1)−1=4f(1) = 2(1)^2 + 3(1) - 1 = 4
This establishes the base value needed for the difference quotient.
2
Expand and simplify f(1+h)f(1+h)
f(1+h)=2(1+h)2+3(1+h)−1=4+7h+2h2f(1+h) = 2(1+h)^2 + 3(1+h) - 1 = 4 + 7h + 2h^2
This gives the value of the function at the incremented point x+hx + h.
3
Form and simplify the difference quotient f(1+h)−f(1)h\frac{f(1+h) - f(1)}{h}
4+7h+2h2−4h=7+2h\frac{4 + 7h + 2h^2 - 4}{h} = 7 + 2h
Dividing by hh eliminates the indeterminate form before taking the limit.
4
Evaluate the limit as h→0h \to 0
f′(1)=lim⁡h→0(7+2h)=7f'(1) = \lim_{h \to 0} (7 + 2h) = 7
Taking h=0h = 0 in the simplified quotient gives the exact rate of change at x=1x = 1.

Key Concept

Differentiation from First Principles
Estimated Time:1m 0s
Question 11Question →

When differentiating the function f(x)=x2+4xf(x) = x^2 + 4x from first principles, what is the simplified expression for the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} before taking the limit as h→0h \to 0?

Show answer & explanation

Answer: 2x + h + 4; 2x + 4 + h; h + 2x + 4

Answer

2x+h+42x + h + 4
Expanding f(x+h)=(x+h)2+4(x+h)=x2+2xh+h2+4x+4hf(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h. Subtracting f(x)=x2+4xf(x) = x^2 + 4x yields 2xh+h2+4h2xh + h^2 + 4h. Dividing every term by hh gives the simplified difference quotient 2x+h+42x + h + 4.

Step-by-Step Solution

1
Evaluate f(x+h)f(x+h) for the function f(x)=x2+4xf(x) = x^2 + 4x
f(x+h)=(x+h)2+4(x+h)=x2+2xh+h2+4x+4hf(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h
Substitute (x+h)(x+h) into every instance of xx in the function and expand the terms.
2
Subtract f(x)f(x) from f(x+h)f(x+h) to find the numerator of the difference quotient
f(x+h)−f(x)=(x2+2xh+h2+4x+4h)−(x2+4x)=2xh+h2+4hf(x+h) - f(x) = (x^2 + 2xh + h^2 + 4x + 4h) - (x^2 + 4x) = 2xh + h^2 + 4h
Subtracting original function terms cancels out x2x^2 and 4x4x.
3
Divide the result by hh
\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 4h}{h} = 2x + h + 4
Factor out hh from the numerator and cancel with the denominator hh.

Key Concept

Difference Quotient Simplification in First Principles Differentiation
Question 12Question →

When differentiating the quadratic function f(x)=5x2−2xf(x) = 5x^2 - 2x from first principles, what is the fully simplified form of the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} before taking the limit as h→0h \to 0?

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Answer: 10x + 5h - 2; 10x - 2 + 5h; 5h + 10x - 2; 10x+5h-2

Answer

10x+5h−210x + 5h - 2
Expanding f(x+h)=5(x+h)2−2(x+h)f(x+h) = 5(x+h)^2 - 2(x+h) yields 5x2+10xh+5h2−2x−2h5x^2 + 10xh + 5h^2 - 2x - 2h. Subtracting f(x)=5x2−2xf(x) = 5x^2 - 2x leaves 10xh+5h2−2h10xh + 5h^2 - 2h. Factoring out hh and dividing by hh gives the simplified difference quotient 10x+5h−210x + 5h - 2.

Step-by-Step Solution

1
Evaluate f(x+h)f(x+h) by expanding 5(x+h)2−2(x+h)5(x+h)^2 - 2(x+h)
f(x+h)=5(x2+2xh+h2)−2x−2h=5x2+10xh+5h2−2x−2hf(x+h) = 5(x^2 + 2xh + h^2) - 2x - 2h = 5x^2 + 10xh + 5h^2 - 2x - 2h
Substitute (x+h)(x+h) into the function definition.
2
Subtract f(x)f(x) from f(x+h)f(x+h)
f(x+h)−f(x)=(5x2+10xh+5h2−2x−2h)−(5x2−2x)=10xh+5h2−2hf(x+h) - f(x) = (5x^2 + 10xh + 5h^2 - 2x - 2h) - (5x^2 - 2x) = 10xh + 5h^2 - 2h
Determine the numerator of the difference quotient by cancelling common terms.
3
Divide the numerator by hh
\frac{f(x+h) - f(x)}{h} = \frac{10xh + 5h^2 - 2h}{h} = 10x + 5h - 2
Simplify the fraction by dividing each term by hh.

Key Concept

Difference quotient in differentiation from first principles
Question 13Question →

Using differentiation from first principles, what is the derivative of the function f(x)=x2−5xf(x) = x^2 - 5x with respect to xx?

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Answer: 2x−52x - 5

Answer

The derivative of f(x)=x2−5xf(x) = x^2 - 5x with respect to xx is 2x−52x - 5.
Evaluating the difference quotient f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h} for f(x)=x2−5xf(x) = x^2 - 5x gives x2+2xh+h2−5x−5h−(x2−5x)h=2x+h−5\frac{x^2+2xh+h^2-5x-5h-(x^2-5x)}{h} = 2x + h - 5. Taking the limit as h→0h \to 0 yields 2x−52x - 5.

Step-by-Step Solution

1
Write down the definition of the derivative from first principles.
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
This is the standard formula for differentiation from first principles.
2
Substitute f(x)=x2−5xf(x) = x^2 - 5x and f(x+h)=(x+h)2−5(x+h)f(x+h) = (x+h)^2 - 5(x+h) into the definition.
\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 5(x+h)] - (x^2 - 5x)}{h}
This sets up the difference quotient for the given polynomial function.
3
Expand the numerator and combine like terms.
\frac{x^2 + 2xh + h^2 - 5x - 5h - x^2 + 5x}{h} = \frac{2xh + h^2 - 5h}{h} = 2x + h - 5
Expanding allows x2x^2 and −5x-5x terms to cancel out, leaving terms containing hh.
4
Evaluate the limit as h→0h \to 0.
\lim_{h \to 0} (2x + h - 5) = 2x - 5
As hh approaches 0, the term hh vanishes, giving the final derivative.

Key Concept

Differentiation from First Principles
Question 14Question →

What is the derivative of the function f(x)=3x2+5xf(x) = 3x^2 + 5x with respect to xx, obtained using differentiation from first principles?

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Answer: 6x+56x + 5

Answer

The derivative of the function is 6x+56x + 5.
Using the first-principles formula f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, we expand f(x+h)=3(x+h)2+5(x+h)=3x2+6xh+3h2+5x+5hf(x+h) = 3(x+h)^2 + 5(x+h) = 3x^2 + 6xh + 3h^2 + 5x + 5h. Subtracting f(x)=3x2+5xf(x) = 3x^2 + 5x yields 6xh+3h2+5h6xh + 3h^2 + 5h. Dividing by hh produces 6x+3h+56x + 3h + 5. Taking the limit as h→0h \to 0 gives 6x+56x + 5.

Step-by-Step Solution

1
Evaluate f(x+h)f(x+h) for f(x)=3x2+5xf(x) = 3x^2 + 5x
f(x+h)=3(x+h)2+5(x+h)=3(x2+2xh+h2)+5x+5h=3x2+6xh+3h2+5x+5hf(x+h) = 3(x+h)^2 + 5(x+h) = 3(x^2 + 2xh + h^2) + 5x + 5h = 3x^2 + 6xh + 3h^2 + 5x + 5h
Substitute (x+h)(x+h) into the original function expression and expand algebraically.
2
Find the difference f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=(3x2+6xh+3h2+5x+5h)−(3x2+5x)=6xh+3h2+5hf(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 5x + 5h) - (3x^2 + 5x) = 6xh + 3h^2 + 5h
Subtract the original function f(x)f(x) to find the net change in output.
3
Divide the difference by hh to set up the difference quotient
\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 + 5h}{h} = 6x + 3h + 5
Cancel out hh from each term in the numerator.
4
Take the limit as h→0h \to 0
f'(x) = \lim_{h \to 0} (6x + 3h + 5) = 6x + 5
Evaluate the expression as hh approaches 0 to determine the instantaneous rate of change.

Key Concept

Differentiation from First Principles
Estimated Time:1m 0s
Question 15Question →

Using differentiation from first principles, what is the derivative dydx\frac{dy}{dx} of the function y=3−2x2y = 3 - 2x^2?

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Answer: −4x-4x

Answer

−4x-4x
Applying the definition of derivative f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} to f(x)=3−2x2f(x) = 3 - 2x^2 gives −4xh−2h2h=−4x−2h\frac{-4xh - 2h^2}{h} = -4x - 2h. Taking the limit as h→0h \to 0 results in −4x-4x.

Step-by-Step Solution

1
Express f(x+h)f(x+h) for the function f(x)=3−2x2f(x) = 3 - 2x^2
f(x+h)=3−2(x+h)2=3−2(x2+2xh+h2)=3−2x2−4xh−2h2f(x+h) = 3 - 2(x+h)^2 = 3 - 2(x^2 + 2xh + h^2) = 3 - 2x^2 - 4xh - 2h^2
Substitute (x+h)(x+h) into the place of xx and expand the squared binomial.
2
Set up the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h}
(3−2x2−4xh−2h2)−(3−2x2)h=−4xh−2h2h\frac{(3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2)}{h} = \frac{-4xh - 2h^2}{h}
Subtract f(x)f(x) from f(x+h)f(x+h) to isolate terms involving increment hh.
3
Divide numerator terms by hh
h(−4x−2h)h=−4x−2h\frac{h(-4x - 2h)}{h} = -4x - 2h
Factor out hh to cancel the denominator.
4
Take the limit as h→0h \to 0
lim⁡h→0(−4x−2h)=−4x\lim_{h \to 0} (-4x - 2h) = -4x
Evaluating the limit gives the exact derivative function dydx\frac{dy}{dx}.

Key Concept

Differentiation from First Principles
Question 16Question →

Using differentiation from first principles, what is the derivative of the function f(x)=4−x2f(x) = 4 - x^2 with respect to xx?

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Answer: −2x-2x

Answer

−2x-2x
Differentiating from first principles involves finding the limit of f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} as h→0h \to 0. For f(x)=4−x2f(x) = 4 - x^2, expanding f(x+h)f(x+h) gives 4−x2−2xh−h24 - x^2 - 2xh - h^2. Subtracting f(x)f(x) yields −2xh−h2-2xh - h^2, and dividing by hh gives −2x−h-2x - h. Taking the limit as h→0h \to 0 leaves −2x-2x.

Step-by-Step Solution

1
Express f(x+h)f(x+h) for f(x)=4−x2f(x) = 4 - x^2
f(x+h)=4−(x+h)2=4−(x2+2xh+h2)=4−x2−2xh−h2f(x+h) = 4 - (x+h)^2 = 4 - (x^2 + 2xh + h^2) = 4 - x^2 - 2xh - h^2
Substitute x+hx+h into the original function definition.
2
Set up the difference f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=(4−x2−2xh−h2)−(4−x2)=−2xh−h2f(x+h) - f(x) = (4 - x^2 - 2xh - h^2) - (4 - x^2) = -2xh - h^2
Subtract f(x)f(x) to find the net change in yy.
3
Divide the difference by hh to form the difference quotient
\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2}{h} = -2x - h
Divide each term in the numerator by hh.
4
Take the limit as h→0h \to 0
f′(x)=lim⁡h→0(−2x−h)=−2xf'(x) = \lim_{h \to 0} (-2x - h) = -2x
Evaluate the derivative by letting hh approach zero.

Key Concept

Differentiation from First Principles
Question 17Question →

Using differentiation from first principles, what is the value of the derivative of the function f(x)=x2+2xf(x) = x^2 + 2x at the point where x=3x = 3?

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Answer: 8

Answer

The derivative of f(x)=x2+2xf(x) = x^2 + 2x evaluated at x=3x = 3 is 8.
Applying first principles to f(x)=x2+2xf(x) = x^2 + 2x yields f′(x)=lim⁡h→0(x+h)2+2(x+h)−(x2+2x)h=lim⁡h→0(2x+h+2)=2x+2f'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - (x^2 + 2x)}{h} = \lim_{h \to 0} (2x + h + 2) = 2x + 2. Evaluating this derivative at x=3x = 3 gives 2(3)+2=82(3) + 2 = 8.

Step-by-Step Solution

1
Apply the definition of differentiation from first principles
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
First principles uses the limit of the difference quotient to compute the instantaneous rate of change.
2
Substitute f(x)=x2+2xf(x) = x^2 + 2x into the difference quotient
[(x+h)2+2(x+h)]−(x2+2x)h=2xh+h2+2hh\frac{[(x+h)^2 + 2(x+h)] - (x^2 + 2x)}{h} = \frac{2xh + h^2 + 2h}{h}
Expanding terms allows cancellation of non-hh terms in the numerator.
3
Simplify the fraction by dividing by hh
2x+h+22x + h + 2
Dividing out hh removes the indeterminate form 00\frac{0}{0}.
4
Evaluate the limit as h→0h \to 0 and substitute x=3x = 3
f′(3)=2(3)+2=8f'(3) = 2(3) + 2 = 8
Setting h=0h = 0 yields the derivative function f′(x)=2x+2f'(x) = 2x + 2, which evaluates to 8 at x=3x = 3.

Key Concept

Differentiation from First Principles
Question 18Question →

By evaluating the limit of the difference quotient lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, determine the value of the derivative of the function f(x)=2x2−4x+5f(x) = 2x^2 - 4x + 5 at the point where x=3x = 3.

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Answer: 8

Answer

The value of the derivative of f(x)=2x2−4x+5f(x) = 2x^2 - 4x + 5 at x=3x = 3 is 8.
Applying first principles, f(x+h)−f(x)=4xh+2h2−4hf(x+h) - f(x) = 4xh + 2h^2 - 4h. Dividing by hh gives 4x+2h−44x + 2h - 4. Taking the limit as h→0h \to 0 produces f′(x)=4x−4f'(x) = 4x - 4. Substituting x=3x = 3 yields 4(3)−4=84(3) - 4 = 8.

Step-by-Step Solution

1
Substitute (x+h)(x+h) into the function f(x)=2x2−4x+5f(x) = 2x^2 - 4x + 5 and expand
f(x+h)=2(x2+2xh+h2)−4x−4h+5=2x2+4xh+2h2−4x−4h+5f(x+h) = 2(x^2 + 2xh + h^2) - 4x - 4h + 5 = 2x^2 + 4xh + 2h^2 - 4x - 4h + 5
Apply algebraic expansion to determine the value of the function at x+hx+h.
2
Form the difference f(x+h)−f(x)f(x+h) - f(x)
f(x+h)−f(x)=4xh+2h2−4hf(x+h) - f(x) = 4xh + 2h^2 - 4h
Subtract the original function terms to leave only terms containing hh.
3
Divide the difference by hh
\frac{f(x+h) - f(x)}{h} = 4x + 2h - 4
Simplify the difference quotient prior to taking the limit.
4
Compute the limit as h→0h \to 0
f′(x)=4x−4f'(x) = 4x - 4
Taking the limit yields the general derivative function f′(x)f'(x).
5
Evaluate f′(x)f'(x) at x=3x = 3
f′(3)=4(3)−4=8f'(3) = 4(3) - 4 = 8
Substitute x=3x = 3 to find the numerical rate of change at the given point.

Key Concept

Differentiation from First Principles
Question 19Question →

When differentiating the function f(x)=x2−3xf(x) = x^2 - 3x from first principles, what is the simplified expression for the difference quotient f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h} for h≠0h \neq 0 before taking the limit as h→0h \to 0?

Show answer & explanation

Answer: 2x + h - 3; 2x - 3 + h; h + 2x - 3; 2x+h-3; 2x-3+h

Answer

2x+h−32x + h - 3
Expanding f(x+h)=(x+h)2−3(x+h)=x2+2xh+h2−3x−3hf(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h and subtracting f(x)=x2−3xf(x) = x^2 - 3x yields 2xh+h2−3h2xh + h^2 - 3h. Factoring and dividing by hh gives the simplified difference quotient 2x+h−32x + h - 3.

Step-by-Step Solution

1
Set up the difference quotient expression
f(x+h)−f(x)h=[(x+h)2−3(x+h)]−(x2−3x)h\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 3(x+h)] - (x^2 - 3x)}{h}
Substitute (x+h)(x+h) into the function definition to evaluate f(x+h)f(x+h).
2
Expand terms in the numerator
(x2+2xh+h2−3x−3h)−x2+3xh\frac{(x^2 + 2xh + h^2 - 3x - 3h) - x^2 + 3x}{h}
Expand (x+h)2=x2+2xh+h2(x+h)^2 = x^2 + 2xh + h^2 and distribute −3-3 across (x+h)(x+h) as well as the negative sign across f(x)f(x).
3
Combine like terms in the numerator
2xh+h2−3hh\frac{2xh + h^2 - 3h}{h}
The terms x2−x2=0x^2 - x^2 = 0 and −3x+3x=0-3x + 3x = 0 cancel out.
4
Divide numerator terms by hh
2x+h−32x + h - 3
Factor out hh from the numerator to get h(2x+h−3)h=2x+h−3\frac{h(2x + h - 3)}{h} = 2x + h - 3 for h≠0h \neq 0.

Key Concept

Difference Quotient in First Principles Differentiation
Estimated Time:1m 30s
Question 20Question →

Using differentiation from first principles, which expression represents the derivative dydx\frac{\mathrm{d}y}{\mathrm{d}x} of the function f(x)=2xf(x) = \frac{2}{x}, where x≠0x \neq 0?

Show answer & explanation

Answer: −2x2-\frac{2}{x^2}

Answer

−2x2-\frac{2}{x^2}
The option stating −2x2-\frac{2}{x^2} is correct because applying the first principles limit formula lim⁡h→02x+h−2xh\lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} gives lim⁡h→0−2hhx(x+h)=lim⁡h→0−2x(x+h)=−2x2\lim_{h \to 0} \frac{-2h}{h x(x+h)} = \lim_{h \to 0} \frac{-2}{x(x+h)} = -\frac{2}{x^2}.

Step-by-Step Solution

1
Set up the difference quotient definition of derivative from first principles.
dydx=lim⁡h→0f(x+h)−f(x)h\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
First principles definition states that the derivative is the limit of the rate of change as hh approaches zero.
2
Substitute f(x)=2xf(x) = \frac{2}{x} and f(x+h)=2x+hf(x+h) = \frac{2}{x+h} into the formula.
\frac{f(x+h) - f(x)}{h} = \frac{\frac{2}{x+h} - \frac{2}{x}}{h}
Replacing the function expressions yields the explicit algebraic fraction.
3
Combine the fractions in the numerator over a common denominator.
\frac{2x - 2(x+h)}{x(x+h) \cdot h} = \frac{2x - 2x - 2h}{h x(x+h)} = \frac{-2h}{h x(x+h)}
Simplifying the numerator yields a factor of hh that can be cancelled with the denominator.
4
Cancel the common factor hh and evaluate the limit as h→0h \to 0.
\lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x(x+0)} = -\frac{2}{x^2}
Dividing out hh allows direct substitution of h=0h = 0 without producing an indeterminate form.

Key Concept

Differentiation from First Principles for Reciprocal Functions
Estimated Time:1m 30s
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