Calculus Alıştırma Soruları — JAMB UTME — Sayfa 5

If $y = e^{3x} \sin(2x)$ , what is $\frac{dy}{dx}$ ?

e^{3x}(3\sin 2x + 2\cos 2x)

e^{3x}(\sin 2x + \cos 2x)

e^{3x}(3\sin 2x - 2\cos 2x)

e^{3x}(2\sin 2x + 3\cos 2x)

Cevabı ve açıklamayı göster

Cevap: $e^{3x}(3\sin 2x + 2\cos 2x)$

Cevap

e^{3x}(3\sin 2x + 2\cos 2x)

Using the product rule

\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

on

y = e^{3x}\sin(2x)

gives

e^{3x} \cdot 2\cos(2x) + \sin(2x) \cdot 3e^{3x}

. Factoring out

e^{3x}

results in

e^{3x}(3\sin 2x + 2\cos 2x)

.

Adım Adım Çözüm

1

Identify the function components for the product rule

Let

u = e^{3x}

and

v = \sin(2x)

.

The function

y

is a product of two differentiable functions

u(x)

and

v(x)

.

2

Differentiate each component using the chain rule

\frac{du}{dx} = 3e^{3x}

and

\frac{dv}{dx} = 2\cos(2x)

.

Applying the chain rule gives

\frac{d}{dx}(e^{3x}) = 3e^{3x}

and

\frac{d}{dx}(\sin 2x) = 2\cos 2x

.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

\frac{dy}{dx} = e^{3x}(2\cos 2x) + \sin(2x)(3e^{3x}) = e^{3x}(3\sin 2x + 2\cos 2x)

.

Combining the products and factoring out the common exponential factor

e^{3x}

yields the final derivative.

Anahtar Kavram

Differentiation of Transcendental Functions using Product Rule and Chain Rule

Tahmini Süre:1m 30s

Soru 82Soru →

If $y = \frac{4x + 1}{x - 2}$ , calculate the numerical value of $\frac{dy}{dx}$ at $x = 3$ .

Cevabı ve açıklamayı göster

Cevap: -9

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 3

is

-9

.

Using the quotient rule

\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

for

y = \frac{4x + 1}{x - 2}

gives

\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{-9}{(x - 2)^2}

. Substituting

x = 3

produces

\frac{-9}{(3 - 2)^2} = -9

.

Adım Adım Çözüm

1

Identify

u(x)

and

v(x)

for the quotient rule formula.

u = 4x + 1

and

v = x - 2

.

The given equation

y = \frac{4x + 1}{x - 2}

is a quotient of two functions of

x

.

2

Calculate the individual derivatives

\frac{du}{dx}

and

\frac{dv}{dx}

.

\frac{du}{dx} = 4

and

\frac{dv}{dx} = 1

.

These derivatives are required components of the quotient rule.

3

Substitute the expressions into the quotient rule formula

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

and simplify.

\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{4x - 8 - 4x - 1}{(x - 2)^2} = \frac{-9}{(x - 2)^2}

.

Simplifying the numerator yields the general derivative function.

4

Evaluate the derivative at

x = 3

.

\frac{dy}{dx}\Big|_{x=3} = \frac{-9}{(3 - 2)^2} = \frac{-9}{1} = -9

.

Substituting

x = 3

gives the requested numerical value.

Anahtar Kavram

Quotient Rule of Differentiation

Soru 83Soru →

Evaluate the limit $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3}$ . What is the numerical value of this limit?

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Cevap: 1

Cevap

The numerical value of the limit is 1.

Substituting x = 3 into the original expression produces the indeterminate form 0/0. Factoring the numerator gives (x - 3)(x - 2). Canceling the common factor (x - 3) yields x - 2. Substituting x = 3 into x - 2 gives 3 - 2 = 1.

Adım Adım Çözüm

1

Check the expression using direct substitution at x = 3.

The substitution yields \frac{3^2 - 5(3) + 6}{3 - 3} = \frac{0}{0}, an indeterminate form.

Direct substitution results in 0/0, indicating that algebraic factorization is needed.

2

Factorize the quadratic polynomial in the numerator.

x^2 - 5x + 6 = (x - 3)(x - 2)

Finding factors of +6 that sum to -5 allows cancellation of the denominator term.

3

Cancel the factor (x - 3) and evaluate the remaining linear expression at x = 3.

\lim_{x \to 3} (x - 2) = 3 - 2 = 1

The factor causing zero in the denominator is removed, leaving a continuous function.

Anahtar Kavram

Evaluating indeterminate limits (0/0) by algebraic factorization

Soru 84Soru →

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x}, & x \neq 0 \\ a^2 - 5, & x = 0 \end{cases}

If

f(x)

is continuous at

x = 0

, where

a > 0

, determine the numerical value of

a

.

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The numerical value of

a

is 3.

For the function to be continuous at

x = 0

, the limit as

x \to 0

must equal the function value

f(0)

. Splitting the trigonometric limit gives

\frac{3}{2} + \frac{5}{2} = 4

. Setting

a^2 - 5 = 4

yields

a^2 = 9

. Because

a > 0

, taking the positive square root gives

a = 3

.

Adım Adım Çözüm

1

Evaluate the limit of the trigonometric expression as

x

approaches 0.

\lim_{x \to 0} \frac{\sin(3x) + \tan(5x)}{2x} = 4

Using standard trigonometric limit principles: \lim_{x \to 0} \frac{\sin(kx)}{x} = k and \lim_{x \to 0} \frac{\tan(kx)}{x} = k.

2

Equate the limit value to

f(0)

to ensure continuity at

x = 0

.

a^2 - 5 = 4

By definition, a function is continuous at x = c if and only if \lim_{x \to c} f(x) = f(c).

3

Solve the resulting quadratic equation for the positive constant

a

.

a = 3

Adding 5 to both sides gives a^2 = 9; taking the principal square root gives a = 3 since a > 0.

Anahtar Kavram

Limits and Continuity of Functions

Soru 85Soru →

If $y = e^{\sin 2x} \ln(\tan x)$ , what is the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$ ?

2e

e

-2e

0

Cevabı ve açıklamayı göster

Cevap: $2e$

Cevap

The derivative evaluated at

x = \frac{\pi}{4}

is

2e

.

Applying the product rule

\frac{dy}{dx} = u'v + uv'

to

y = e^{\sin 2x} \ln(\tan x)

gives

u' = 2\cos(2x)e^{\sin 2x}

and

v' = 2\csc(2x)

. At

x = \frac{\pi}{4}

,

\cos(2x) = \cos(\pi/2) = 0

, which vanishes the first term. The second term evaluates to

e^{\sin(\pi/2)} \cdot 2\csc(\pi/2) = e^1 \cdot 2(1) = 2e

.

Adım Adım Çözüm

1

Identify the product rule structure for

y = u \cdot v

, where

u = e^{\sin 2x}

and

v = \ln(\tan x)

.

The product rule states

\frac{dy}{dx} = u'v + uv'

.

The given function is a product of an exponential function and a logarithmic function.

2

Differentiate

u = e^{\sin 2x}

using the chain rule.

\frac{du}{dx} = \frac{d}{dx}(\sin 2x) \cdot e^{\sin 2x} = 2\cos(2x) e^{\sin 2x}

.

The inner function is

\sin 2x

, whose derivative is

2\cos 2x

.

3

Differentiate

v = \ln(\tan x)

using the chain rule.

\frac{dv}{dx} = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2\csc(2x)

.

Simplifying

\frac{\sec^2 x}{\tan x}

gives

\frac{2}{\sin 2x} = 2\csc(2x)

.

4

Combine terms into the product rule expression.

\frac{dy}{dx} = [2\cos(2x) e^{\sin 2x} \ln(\tan x)] + [e^{\sin 2x} \cdot 2\csc(2x)]

.

Substitute

u, u', v, v'

into

\frac{dy}{dx} = u'v + uv'

.

5

Evaluate the derivative at

x = \frac{\pi}{4}

.

At

x = \frac{\pi}{4}

,

2x = \frac{\pi}{2}

. Thus,

\cos(\pi/2) = 0

,

\tan(\pi/4) = 1 \implies \ln(1) = 0

,

\sin(\pi/2) = 1

, and

\csc(\pi/2) = 1

. The first term becomes

0

, and the second term becomes

e^{1} \cdot 2(1) = 2e

. Thus,

\frac{dy}{dx} = 2e

.

Substitute special angle values into the expression to compute the numerical result.

Anahtar Kavram

Product Rule and Chain Rule for Transcendental Functions

Soru 86Soru →

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

6x^2 + 6xh + 2h^2 - 5

6x^2 + 6xh + 2h^2 + 5

2x^2 + 2xh + 2h^2 - 5

6x^2 - 5

Cevabı ve açıklamayı göster

Cevap: $6x^2 + 6xh + 2h^2 - 5$

Cevap

The simplified difference quotient before evaluating the limit is

6x^2 + 6xh + 2h^2 - 5

.

Expanding

f(x+h) = 2(x+h)^3 - 5(x+h)

using the binomial expansion yields

2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h

. Subtracting

f(x) = 2x^3 - 5x

leaves

6x^2h + 6xh^2 + 2h^3 - 5h

. Dividing each term by

h

gives the correct simplified difference quotient

6x^2 + 6xh + 2h^2 - 5

.

Adım Adım Çözüm

1

Express

f(x+h)

for the function

f(x) = 2x^3 - 5x

f(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)

into

f(x)

and expand the binomial terms completely.

2

Subtract

f(x)

from

f(x+h)

to find the numerator

f(x+h) - f(x)

f(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

2x^3

and

-5x

.

3

Divide the numerator by

h

to form the difference quotient

\frac{6x^2h + 6xh^2 + 2h^3 - 5h}{h} = 6x^2 + 6xh + 2h^2 - 5

Factor out

h

from each term in the numerator and divide by

h

.

Anahtar Kavram

Differentiation from First Principles

Soru 87Soru →

A normal line is drawn to the curve $y = \frac{x^2 - 5}{x - 3}$ at the point where $x = 2$ . What is the $x$ -intercept of this normal line?

Cevabı ve açıklamayı göster

Cevap: -1

Cevap

The

x

-intercept of the normal line is

-1

.

Substituting

x = 2

into the curve yields the point

(2, 1)

. Differentiation via the quotient rule gives

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

. At

x = 2

, the tangent slope is

-3

, making the normal slope

\frac{1}{3}

. The line equation

y - 1 = \frac{1}{3}(x - 2)

simplifies to

x - 3y + 1 = 0

. Setting

y = 0

gives

x = -1

.

Adım Adım Çözüm

1

Calculate the y-coordinate of the point of tangency

y = 1

Substitute

x = 2

into the curve equation

y = \frac{x^2 - 5}{x - 3}

to get the point

(2, 1)

.

2

Differentiate the function using the quotient rule

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

Applying

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

to

u = x^2 - 5

and

v = x - 3

.

3

Find the tangent gradient at

x = 2

m_t = -3

Evaluate

\frac{dy}{dx}

at

x = 2

to obtain

m_t = \frac{2^2 - 6(2) + 5}{(2-3)^2} = -3

.

4

Calculate the gradient of the normal line

m_n = \frac{1}{3}

The normal line is perpendicular to the tangent line, so

m_n = -\frac{1}{m_t} = \frac{1}{3}

.

5

Derive the line equation for the normal

x - 3y + 1 = 0

Use point-slope form

y - y_1 = m_n(x - x_1)

with

(2, 1)

and

m_n = \frac{1}{3}

.

6

Find the x-intercept

x = -1

Set

y = 0

in the normal equation

x - 3(0) + 1 = 0

, giving

x = -1

.

Anahtar Kavram

Equation of normal line to a rational curve and finding its axis intercepts

Soru 88Soru →

A curve has the equation $y = ax^3 + bx^2 + 12x + 1$ , where $a$ and $b$ are constants. If the curve has stationary points at $x = 1$ and $x = 2$ , what is the value of $a$ ?

2

6

-2

-6

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The value of

a

is

2

.

To find

a

, take the first derivative of the curve, yielding

\frac{dy}{dx} = 3ax^2 + 2bx + 12

. Setting this to zero gives a quadratic equation with roots

x = 1

and

x = 2

. The product of roots for a quadratic equation

Ax^2 + Bx + C = 0

is

\frac{C}{A}

. Therefore,

1 \times 2 = \frac{12}{3a}

, which simplifies to

2 = \frac{4}{a}

, giving

a = 2

.

Adım Adım Çözüm

1

Find the derivative of the given curve with respect to

x

.

\frac{dy}{dx} = 3ax^2 + 2bx + 12

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the derivative to zero and substitute the stationary point locations

x = 1

and

x = 2

.

The roots of the quadratic equation

3ax^2 + 2bx + 12 = 0

are

x_1 = 1

and

x_2 = 2

.

Since stationary points are given at

x=1

and

x=2

, these values satisfy

\frac{dy}{dx} = 0

.

3

Apply the product of roots formula for a quadratic equation

Ax^2 + Bx + C = 0

, where

x_1 \cdot x_2 = \frac{C}{A}

.

1 \times 2 = \frac{12}{3a} \implies 2 = \frac{4}{a}

Equating the product of roots

1 \times 2 = 2

to

\frac{12}{3a}

isolates parameter

a

.

4

Solve for

a

.

a = 2

Multiplying both sides by

a

gives

2a = 4

, so

a = 2

.

Anahtar Kavram

Stationary points occur where

\frac{dy}{dx} = 0

. For a cubic curve, the derivative is a quadratic equation whose roots correspond to the

x

-coordinates of the stationary points.

Tahmini Süre:2m 0s

Soru 89Soru →

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

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The numerical value of the derivative at

x = 2

is 12.

Applying the first principles limit formula

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

to

f(x) = 2x^3 - 3x^2 + 4

yields

f'(x) = 6x^2 - 6x

. Evaluating this derivative function at

x = 2

yields

6(4) - 12 = 12

.

Adım Adım Çözüm

1

Set up the difference quotient using the definition of differentiation from first principles

f'(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

h

approaches zero.

2

Expand

(x+h)^3

and

(x+h)^2

and subtract

f(x)

f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 6xh - 3h^2

Expanding the terms allows cancellation of all terms not containing

h

.

3

Divide by

h

and evaluate the limit as

h \to 0

f'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 6x - 3h) = 6x^2 - 6x

Dividing by

h

eliminates the indeterminate form

\frac{0}{0}

, allowing direct substitution of

h = 0

.

4

Substitute

x = 2

into the derivative function

f'(x)

f'(2) = 6(2)^2 - 6(2) = 24 - 12 = 12

Evaluating at the given point gives the slope of the tangent line at

x = 2

.

Anahtar Kavram

Differentiation from first principles using limit of difference quotient

Soru 90Soru →

What is the derivative of the function $f(x) = \frac{\sin(2x)}{(1 + x^2)^2}$ with respect to $x$ ?

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

\frac{(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}

\frac{2[(1 + x^2)\cos(2x) - \sin(2x)]}{(1 + x^2)^3}

\frac{2[(1 + x^2)\cos(2x) + 2x\sin(2x)]}{(1 + x^2)^3}

Cevabı ve açıklamayı göster

Cevap: $\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}$

Cevap

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Applying the quotient rule

\frac{u'v - uv'}{v^2}

with

u = \sin(2x)

and

v = (1 + x^2)^2

yields

u' = 2\cos(2x)

and

v' = 4x(1 + x^2)

. Substituting these into the formula gives

\frac{2\cos(2x)(1+x^2)^2 - 4x\sin(2x)(1+x^2)}{(1+x^2)^4}

. Factoring out

(1+x^2)

from numerator and denominator reduces the expression to

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

.

Adım Adım Çözüm

1

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = \sin(2x)

and

v(x) = (1 + x^2)^2

The quotient rule states that

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

.

2

Find the derivatives

u'(x)

and

v'(x)

using the chain rule.

u'(x) = 2\cos(2x)

and

v'(x) = 2(1 + x^2) \cdot (2x) = 4x(1 + x^2)

By the chain rule,

\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2

and

\frac{d}{dx}(1 + x^2)^2 = 2(1 + x^2) \cdot 2x

.

3

Substitute

u, u', v, v'

into the quotient rule formula.

f'(x) = \frac{2\cos(2x)(1 + x^2)^2 - \sin(2x) \cdot 4x(1 + x^2)}{((1 + x^2)^2)^2}

Applying

\frac{u'v - uv'}{v^2}

directly.

4

Factor out

(1 + x^2)

from the numerator and simplify the denominator.

f'(x) = \frac{(1 + x^2)[2\cos(2x)(1 + x^2) - 4x\sin(2x)]}{(1 + x^2)^4} = \frac{2(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}

Canceling the common factor

(1 + x^2)

from numerator and denominator.

5

Factor out the constant

2

from the numerator.

f'(x) = \frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Putting the derivative into simplified factored form.

Anahtar Kavram

Combining Quotient Rule and Chain Rule for Composite Functions

Soru 91Soru →

What is the gradient of the tangent line to the curve $y = 4x^2 - 7x + 2$ at the point where $x = 3$ ?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

The gradient of the tangent line to the curve at

x = 3

is 17.

Differentiating

y = 4x^2 - 7x + 2

gives

\frac{dy}{dx} = 8x - 7

. Substituting

x = 3

yields

8(3) - 7 = 17

.

Adım Adım Çözüm

1

Find the derivative of the curve's equation

\frac{dy}{dx} = 8x - 7

The gradient of the curve at any point is given by its first derivative with respect to x.

2

Evaluate the derivative at x = 3

m = 8(3) - 7 = 17

Substituting the given x-coordinate into the derivative gives the specific slope of the tangent line at that point.

Anahtar Kavram

The gradient of the tangent to a curve

y = f(x)

at

x = a

is the value of the first derivative

f'(a)

.

Soru 92Soru →

If $y = (x^2 - 2x + 2)^3$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 2$ ?

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

24

Applying the chain rule gives

\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2)

. Evaluating this derivative at

x = 2

yields

3(2)^2(2) = 24

.

Adım Adım Çözüm

1

Identify the composite structure and state the chain rule formula.

Let

u = x^2 - 2x + 2

, so

y = u^3

. By the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

The function is an expression raised to a power, requiring the chain rule.

2

Differentiate the outer function with respect to

u

and the inner function with respect to

x

.

\frac{dy}{du} = 3(x^2 - 2x + 2)^2 and \frac{du}{dx} = 2x - 2.

Apply the power rule to both the outer expression and inner polynomial.

3

Multiply the two derivatives to express

\frac{dy}{dx}

.

\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2).

Product of outer and inner derivatives per the chain rule.

4

Evaluate the expression at

x = 2

.

\frac{dy}{dx}\Big|_{x=2} = 3(4 - 4 + 2)^2(4 - 2) = 3(2)^2(2) = 24.

Substitute

x = 2

into the derivative expression and simplify.

Anahtar Kavram

Chain Rule of Differentiation

Soru 93Soru →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{1 - \cos(6x)}{x \tan(3x)}, & x \neq 0 \\ p + 4, & x = 0 \end{cases}

If

f(x)

is continuous at

x = 0

, what is the value of the constant

p

?

6

2

-1

-2

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the constant

p

is

2

.

For the piecewise function to be continuous at

x = 0

, the limit

\lim_{x \to 0} f(x)

must exist and equal

f(0) = p + 4

. By applying the double-angle identity

1 - \cos(6x) = 2\sin^2(3x)

and using the standard limit

\lim_{u \to 0} \frac{\sin u}{u} = 1

, the limit evaluates to

6

. Equating

p + 4 = 6

yields

p = 2

.

Adım Adım Çözüm

1

State the continuity condition at

x = 0

\lim_{x \to 0} f(x) = f(0) = p + 4

For a function to be continuous at a point, its limit at that point must equal the defined function value.

2

Apply the trigonometric identity

1 - \cos(6x) = 2\sin^2(3x)

\frac{1 - \cos(6x)}{x \tan(3x)} = \frac{2\sin^2(3x)}{x \cdot \frac{\sin(3x)}{\cos(3x)}} = 2\cos(3x) \cdot \frac{\sin(3x)}{x}

Rewriting

\tan(3x)

as

\frac{\sin(3x)}{\cos(3x)}

allows cancellation of one

\sin(3x)

factor.

3

Evaluate the limit as

x \to 0

\lim_{x \to 0} \left[2\cos(3x) \cdot 3 \cdot \frac{\sin(3x)}{3x}\right] = 2(1)(3)(1) = 6

Using the standard trigonometric limit \lim_{u \to 0} \frac{\sin(u)}{u} = 1 and \cos(0) = 1.

4

Equate the limit to

f(0)

and solve for

p

p + 4 = 6 \implies p = 2

Subtracting 4 from both sides isolates the parameter

p

.

Anahtar Kavram

Continuity of a Piecewise Function using Trigonometric Limits

Tahmini Süre:2m 0s

Soru 94Soru →

If $y = \frac{(2x^2 + 1)^3}{4x - 1}$ , determine the numerical value of $\frac{dy}{dx}$ at $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

The numerical value of the derivative

\frac{dy}{dx}

at

x = 1

is 24.

Evaluating the derivative requires combined application of the quotient rule and chain rule. Differentiating

u(x) = (2x^2 + 1)^3

via the chain rule yields

u'(x) = 12x(2x^2 + 1)^2

. At

x = 1

,

u(1) = 27

,

u'(1) = 108

,

v(1) = 3

, and

v'(1) = 4

. Substituting these into the quotient rule

\frac{u'v - uv'}{v^2}

produces

\frac{324 - 108}{9} = 24

.

Adım Adım Çözüm

1

Decompose the function for the Quotient Rule

Let

u(x) = (2x^2 + 1)^3

and

v(x) = 4x - 1

.

The given expression is a quotient of two differentiable functions.

2

Differentiate the numerator using the Chain Rule

u'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

The chain rule states that

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

.

3

Differentiate the denominator and evaluate all components at

x = 1

v'(1) = 4

,

v(1) = 3

,

u(1) = 27

, and

u'(1) = 108

Evaluating each term individually reduces algebraic complexity prior to applying the quotient rule.

4

Apply the Quotient Rule formula and evaluate

\frac{dy}{dx}\Big|_{x=1} = \frac{(108)(3) - (27)(4)}{3^2} = \frac{324 - 108}{9} = 24

The quotient rule formula is

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

.

Anahtar Kavram

Combination of Quotient Rule and Chain Rule

Tahmini Süre:1m 30s

Soru 95Soru →

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

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of the derivative at

x = 1

is 7.

Using first principles, the derivative is defined as

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

. Evaluating at

x = 1

, we find

f(1) = 4

and

f(1+h) = 4 + 7h + 2h^2

. Subtracting

f(1)

leaves

7h + 2h^2

, and dividing by

h

yields

7 + 2h

. Taking the limit as

h \to 0

yields the final result of 7.

Adım Adım Çözüm

1

Calculate the value of the function at

x = 1

f(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) = 2(1+h)^2 + 3(1+h) - 1 = 4 + 7h + 2h^2

This gives the value of the function at the incremented point

x + h

.

3

Form and simplify the difference quotient

\frac{f(1+h) - f(1)}{h}

\frac{4 + 7h + 2h^2 - 4}{h} = 7 + 2h

Dividing by

h

eliminates the indeterminate form before taking the limit.

4

Evaluate the limit as

h \to 0

f'(1) = \lim_{h \to 0} (7 + 2h) = 7

Taking

h = 0

in the simplified quotient gives the exact rate of change at

x = 1

.

Anahtar Kavram

Differentiation from First Principles

Tahmini Süre:1m 0s

Soru 96Soru →

If $y = (3x - 1)^4$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

96

Applying the chain rule

\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx}

to

y = (3x - 1)^4

gives

\frac{dy}{dx} = 4(3x - 1)^3 \cdot 3 = 12(3x - 1)^3

. Evaluating this expression at

x = 1

yields

12(3(1) - 1)^3 = 12(2)^3 = 96

.

Adım Adım Çözüm

1

Differentiate

y = (3x - 1)^4

with respect to

x

using the chain rule.

\frac{dy}{dx} = 12(3x - 1)^3

Differentiate the outer power function to get

4(3x - 1)^3

and multiply by the derivative of the inner linear function

3x - 1

, which is 3.

2

Substitute

x = 1

into the derivative function.

96

Evaluating

12(3(1) - 1)^3

gives

12(2)^3 = 12 \times 8 = 96

.

Anahtar Kavram

Chain Rule of Differentiation

Soru 97Soru →

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

Cevabı ve açıklamayı göster

Cevap: 2x + h + 4; 2x + 4 + h; h + 2x + 4

Cevap

2x + h + 4

Expanding

f(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h

. Subtracting

f(x) = x^2 + 4x

yields

2xh + h^2 + 4h

. Dividing every term by

h

gives the simplified difference quotient

2x + h + 4

.

Adım Adım Çözüm

1

Evaluate

f(x+h)

for the function

f(x) = x^2 + 4x

f(x+h) = (x+h)^2 + 4(x+h) = x^2 + 2xh + h^2 + 4x + 4h

Substitute

(x+h)

into every instance of

x

in the function and expand the terms.

2

Subtract

f(x)

from

f(x+h)

to find the numerator of the difference quotient

f(x+h) - f(x) = (x^2 + 2xh + h^2 + 4x + 4h) - (x^2 + 4x) = 2xh + h^2 + 4h

Subtracting original function terms cancels out

x^2

and

4x

.

3

Divide the result by

h

\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 4h}{h} = 2x + h + 4

Factor out

h

from the numerator and cancel with the denominator

h

.

Anahtar Kavram

Difference Quotient Simplification in First Principles Differentiation

Soru 98Soru →

Find the $x$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The

x

-intercept of the normal line is

2

.

At

x = 1

, the point on the curve

y = x^3 - 3x^2 + 4x - 1

is

(1, 1)

. The derivative is

\frac{dy}{dx} = 3x^2 - 6x + 4

, which equals

1

at

x = 1

. The normal gradient is therefore

-1

. The normal line equation is

y - 1 = -1(x - 1)

, which simplifies to

y = -x + 2

. Setting

y = 0

gives

x = 2

.

Adım Adım Çözüm

1

Find the y-coordinate at x = 1

y = 1, so the point on the curve is (1, 1)

The point of contact is needed to construct the equation of the normal line.

2

Differentiate the function with respect to x

dy/dx = 3x^2 - 6x + 4

The derivative gives the expression for the gradient of the tangent to the curve.

3

Find the gradient of the tangent and normal at x = 1

Gradient of tangent m_t = 1; gradient of normal m_n = -1

The normal line is perpendicular to the tangent line, so m_n = -1 / m_t.

4

Determine the equation of the normal line

y - 1 = -1(x - 1) => y = -x + 2

Applying the straight-line equation y - y_1 = m(x - x_1).

5

Find the x-intercept of the normal line

x = 2

Setting y = 0 in the normal line equation yields the x-intercept.

Anahtar Kavram

Equation and axis intercepts of a normal line to a curve

Soru 99Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

10x + 5h - 2

Expanding

f(x+h) = 5(x+h)^2 - 2(x+h)

yields

5x^2 + 10xh + 5h^2 - 2x - 2h

. Subtracting

f(x) = 5x^2 - 2x

leaves

10xh + 5h^2 - 2h

. Factoring out

h

and dividing by

h

gives the simplified difference quotient

10x + 5h - 2

.

Adım Adım Çözüm

1

Evaluate

f(x+h)

by expanding

5(x+h)^2 - 2(x+h)

f(x+h) = 5(x^2 + 2xh + h^2) - 2x - 2h = 5x^2 + 10xh + 5h^2 - 2x - 2h

Substitute

(x+h)

into the function definition.

2

Subtract

f(x)

from

f(x+h)

f(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

h

\frac{f(x+h) - f(x)}{h} = \frac{10xh + 5h^2 - 2h}{h} = 10x + 5h - 2

Simplify the fraction by dividing each term by

h

.

Anahtar Kavram

Difference quotient in differentiation from first principles

Soru 100Soru →

What is the gradient of the normal to the curve $y = x^2 - 4x + 5$ at the point where $x = 3$ ?

-\frac{1}{2}

2

\frac{1}{2}

-2

Cevabı ve açıklamayı göster

Cevap: $-\frac{1}{2}$

Cevap

The gradient of the normal to the curve is

-\frac{1}{2}

.

Differentiating

y = x^2 - 4x + 5

gives

\frac{dy}{dx} = 2x - 4

. Substituting

x = 3

yields a tangent gradient of

2

. Since the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal,

-\frac{1}{2}

.

Adım Adım Çözüm

1

Differentiate the equation of the curve to determine the gradient function.

\frac{dy}{dx} = 2x - 4

The first derivative represents the gradient of the tangent to the curve at any point

x

.

2

Evaluate the derivative at

x = 3

to find the tangent gradient

m_t

.

m_t = 2(3) - 4 = 2

Substituting

x = 3

gives the slope of the tangent line at the given point.

3

Calculate the gradient of the normal

m_n

using

m_n = -\frac{1}{m_t}

.

m_n = -\frac{1}{2}

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent's gradient.

Anahtar Kavram

Gradient of a Normal Line

Calculus

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