Rules of Differentiation (Product, Quotient, and Chain Rules) Alıştırma Soruları — JAMB UTME

Given the function $y = \frac{(x^2 + 1)^3}{3x - 5}$ , calculate the value of $\frac{dy}{dx}$ at $x = 2$ .

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Cevap: -75

Cevap

The value of

\frac{dy}{dx}

at

x = 2

is

-75

.

Applying the Quotient Rule

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

along with the Chain Rule to differentiate

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

yields

u'(x) = 6x(x^2+1)^2

. Evaluating at

x=2

gives

u(2)=125

,

u'(2)=300

,

v(2)=1

, and

v'(2)=3

, leading to

\frac{300(1) - 125(3)}{1^2} = -75

.

Adım Adım Çözüm

1

Set up the Quotient Rule framework

Let

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

and

v(x) = 3x - 5

, so that

y = \frac{u(x)}{v(x)}

and

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

.

The function is expressed as a quotient of two differentiable terms.

2

Differentiate the numerator using the Chain Rule

u'(x) = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 6x(x^2 + 1)^2

.

The numerator is a composite function requiring the inner derivative derivative of

x^2+1

to be multiplied.

3

Differentiate the denominator

v'(x) = 3

.

The derivative of a linear function

3x - 5

with respect to

x

is its coefficient 3.

4

Evaluate

u(x)

,

u'(x)

,

v(x)

, and

v'(x)

at

x = 2

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

,

v'(2) = 3

.

Substituting

x = 2

into each evaluated component simplifies the numerical calculation.

5

Substitute numerical values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(3)}{1^2} = \frac{300 - 375}{1} = -75

.

Completing the arithmetic calculation yields the final numerical derivative value.

Anahtar Kavram

Combining the Quotient Rule and Chain Rule for composite fractional functions

Tahmini Süre:1m 30s

Soru 2Soru →

Given the function $y = x^2 \sqrt{3x^2 + 4}$ , find the numerical value of $\frac{dy}{dx}$ at $x = 2$ .

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Cevap: 22

Cevap

The numerical value of the derivative at

x = 2

is 22.

Applying both the Product Rule and Chain Rule correctly yields

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

. Evaluating this expression at

x = 2

gives

16 + 6 = 22

.

Adım Adım Çözüm

1

Decompose the function into a product of two functions

Let

u(x) = x^2

and

v(x) = \sqrt{3x^2 + 4} = (3x^2 + 4)^{1/2}

.

The function

y

is expressed as the product of

u(x)

and

v(x)

, requiring the Product Rule for differentiation.

2

Find the derivative of the inner square root function using the Chain Rule

v'(x) = \frac{1}{2}(3x^2 + 4)^{-1/2} \cdot 6x = \frac{3x}{\sqrt{3x^2 + 4}}

.

The Chain Rule states that

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

.

3

Combine derivatives using the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2x\sqrt{3x^2 + 4} + \frac{3x^3}{\sqrt{3x^2 + 4}}

.

The Product Rule formula is

\frac{d}{dx}[u \cdot v] = u'v + uv'

.

4

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)\sqrt{3(2)^2 + 4} + \frac{3(2)^3}{\sqrt{3(2)^2 + 4}} = 4(4) + \frac{24}{4} = 16 + 6 = 22

.

Substituting

x = 2

gives the specific slope of the tangent line to the curve at that point.

Anahtar Kavram

Combining the Product Rule and Chain Rule to evaluate derivatives of composite product functions

Soru 3Soru →

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

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Cevap: 54

Cevap

54

Applying the chain rule gives

\frac{dy}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2

. Evaluating at

x = 1

gives

6(3)^2 = 54

.

Adım Adım Çözüm

1

Differentiate

y = (2x + 1)^3

using the chain rule.

\frac{dy}{dx} = 6(2x + 1)^2

According to the chain rule,

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

, where

u = 2x + 1

and

\frac{du}{dx} = 2

.

2

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = 54

Substituting

x = 1

into

6(2x + 1)^2

yields

6(3)^2 = 54

.

Anahtar Kavram

Chain Rule for Differentiation

Soru 4Soru →

If $y = (2x - 1)^3 (x^2 + 3)^2$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

112

56

80

104

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Cevap: $112$

Cevap

112

Applying the product rule

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

along with the chain rule gives

u'(1) = 6

,

v(1) = 16

,

u(1) = 1

, and

v'(1) = 16

. Evaluating

u'(1)v(1) + u(1)v'(1)

yields

6 \times 16 + 1 \times 16 = 96 + 16 = 112

.

Adım Adım Çözüm

1

Decompose the function into two components for the Product Rule

Let

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

and

v(x) = (x^2 + 3)^2

, so

y = u(x) v(x)

.

The function is a product of two composite expressions.

2

Differentiate each term using the Chain Rule

u'(x) = 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 6(2x - 1)^2

v'(x) = 2(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) = 4x(x^2 + 3)

Applying the Chain Rule requires differentiating the outer function and multiplying by the derivative of the inner function.

3

Apply the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 6(2x - 1)^2 (x^2 + 3)^2 + 4x(2x - 1)^3 (x^2 + 3)

The derivative of a product

u \cdot v

is

u'v + uv'

.

4

Evaluate the derivative at

x = 1

u(1) = (2(1) - 1)^3 = 1

u'(1) = 6(2(1) - 1)^2 = 6

v(1) = (1^2 + 3)^2 = 16

v'(1) = 4(1)(1^2 + 3) = 16

\left.\frac{dy}{dx}\right|_{x=1} = (6)(16) + (1)(16) = 96 + 16 = 112

Substitute

x = 1

into all expressions to find the numerical derivative.

Anahtar Kavram

Combined Product Rule and Chain Rule of Differentiation

Tahmini Süre:2m 0s

Soru 5Soru →

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

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Cevap: 34

Cevap

The value of

\frac{dy}{dx}

at

x = 2

is

34

.

Applying the product rule

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

alongside the chain rule gives

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

. Evaluating at

x = 2

yields

4(1) + 30(1) = 34

.

Adım Adım Çözüm

1

Identify component functions for the product rule

Let

u(x) = x^2 + 1

and

v(x) = (2x - 3)^3

.

The function

y

is a product of two differentiable functions.

2

Differentiate each component function

u'(x) = 2x

and

v'(x) = 3(2x - 3)^2 \cdot 2 = 6(2x - 3)^2

.

The power rule gives

u'(x)

and the chain rule gives

v'(x)

by multiplying by the derivative of the inner function

(2x - 3)

.

3

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

.

To find the general derivative of a product of functions.

4

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34

.

To calculate the specific numerical value of the derivative at

x = 2

.

Anahtar Kavram

Product and Chain Rules of Differentiation

Tahmini Süre:1m 30s

Soru 6Soru →

Given that $y = \frac{3x - 1}{(x + 2)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

\frac{5}{27}

\frac{13}{9}

\frac{25}{81}

\frac{1}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{5}{27}$

Cevap

The value of

\frac{dy}{dx}

at

x = 1

is

\frac{5}{27}

.

Applying the quotient rule

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

with

u = 3x - 1

and

v = (x + 2)^2

yields

\frac{dy}{dx} = \frac{8 - 3x}{(x + 2)^3}

. Substituting

x = 1

results in

\frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{27}

.

Adım Adım Çözüm

1

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

Let

u = 3x - 1

and

v = (x + 2)^2

.

The given function

y = \frac{u}{v}

requires the quotient rule

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

.

2

Differentiate

u(x)

and

v(x)

with respect to

x

.

u' = 3

and, using the chain rule,

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

.

The derivative of the inner term

(x + 2)

is

1

, giving

v' = 2(x + 2)

.

3

Substitute

u, v, u', v'

into the quotient rule formula and simplify.

\frac{dy}{dx} = \frac{3(x + 2)^2 - (3x - 1) \cdot 2(x + 2)}{(x + 2)^4} = \frac{(x + 2)[3(x + 2) - 2(3x - 1)]}{(x + 2)^4} = \frac{3x + 6 - 6x + 2}{(x + 2)^3} = \frac{8 - 3x}{(x + 2)^3}

.

Factoring out

(x + 2)

simplifies the algebraic expression.

4

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = \frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{3^3} = \frac{5}{27}

.

Substituting

x = 1

gives the final numerical derivative value.

Anahtar Kavram

Quotient and Chain Rules of Differentiation

Soru 7Soru →

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

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Cevap: 1

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is 1.

Applying the quotient rule

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

alongside the chain rule for the denominator yields

\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}

. Substituting

x = 1

evaluates to

\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1

.

Adım Adım Çözüm

1

Identify the numerator and denominator functions

Let

u(x) = 2x - 3

and

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

.

The given function is structured as a quotient

y = \frac{u}{v}

, requiring the quotient rule.

2

Find the derivatives

u'(x)

and

v'(x)

u'(x) = 2

and

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

.

Differentiating

u(x)

follows standard polynomial rules;

v(x)

requires the chain rule.

3

Evaluate

u(1), u'(1), v(1),

and

v'(1)

at

x = 1

u(1) = -1

,

u'(1) = 2

,

v(1) = 4

, and

v'(1) = 8

.

Evaluating components before substitution simplifies the arithmetic.

4

Apply the quotient rule formula

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

at

x = 1

\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.

Substitute the calculated component values into the quotient rule formula.

Anahtar Kavram

Quotient Rule and Chain Rule of Differentiation

Tahmini Süre:1m 30s

Soru 8Soru →

Given the function $y = \frac{x^2 + 2x}{(3x - 1)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

-\frac{5}{4}

\frac{1}{4}

\frac{13}{4}

\frac{1}{3}

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Cevap: $-\frac{5}{4}$

Cevap

-\frac{5}{4}

Applying the quotient rule

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

with

u = x^2 + 2x

(

u' = 2x + 2

) and

v = (3x - 1)^2

(

v' = 6(3x - 1)

) yields

\frac{dy}{dx} = \frac{16 - 36}{16} = -\frac{5}{4}

at

x = 1

.

Adım Adım Çözüm

1

Identify the numerator and denominator functions for the quotient rule

y = \frac{u}{v}

.

u = x^2 + 2x

and

v = (3x - 1)^2

.

The function is structured as a quotient of two algebraic expressions.

2

Differentiate

u

and

v

with respect to

x

.

\frac{du}{dx} = 2x + 2

and

\frac{dv}{dx} = 2(3x - 1) \cdot 3 = 6(3x - 1)

.

Use the power rule for

u

and the chain rule for

v

.

3

Apply the quotient rule formula

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

.

\frac{dy}{dx} = \frac{(3x - 1)^2 (2x + 2) - (x^2 + 2x) \cdot 6(3x - 1)}{(3x - 1)^4}

.

The quotient rule formula combines the expressions and their derivatives.

4

Substitute

x = 1

into the derivative expression and simplify.

\frac{dy}{dx} = \frac{(2)^2 (4) - (3) \cdot 6(2)}{(2)^4} = \frac{16 - 36}{16} = -\frac{20}{16} = -\frac{5}{4}

.

Evaluating at

x = 1

yields the numerical derivative value.

Anahtar Kavram

Quotient Rule and Chain Rule of Differentiation

Soru 9Soru →

Given the function $y = \frac{(x^2 + 1)^3}{2x - 3}$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 2$ ?

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Cevap: 50

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 2

is 50.

Applying the Quotient Rule

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

along with the Chain Rule to find

u'(x) = 6x(x^2 + 1)^2

, and evaluating all components at

x = 2

gives

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

, and

v'(2) = 2

. Substituting these values yields

\frac{(300)(1) - (125)(2)}{1^2} = 50

.

Adım Adım Çözüm

1

Set up the Quotient Rule components

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

and

v(x) = 2x - 3

The given expression is a quotient of two functions requiring

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

.

2

Differentiate

u(x)

using the Chain Rule and

v(x)

using basic power rules

u'(x) = 6x(x^2 + 1)^2

and

v'(x) = 2

Differentiating the outer power 3 gives

3(x^2 + 1)^2

, and multiplying by the derivative of the inner function

(2x)

gives

6x(x^2 + 1)^2

.

3

Evaluate

u(2)

,

u'(2)

,

v(2)

, and

v'(2)

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

, and

v'(2) = 2

Substituting

x = 2

into each function and derivative simplifies calculation of the overall derivative.

4

Substitute values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(2)}{(1)^2} = 50

Evaluating

\frac{u'(2)v(2) - u(2)v'(2)}{[v(2)]^2}

yields the exact numerical result.

Anahtar Kavram

Combined Application of Quotient Rule and Chain Rule

Soru 10Soru →

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

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Cevap: 32

Cevap

The numerical value of

\frac{dy}{dx}

evaluated at

x = 1

is

32

.

Using the product rule together with the chain rule for composite functions, we find

u'(x) = 6(2x - 1)^2

and

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

. Evaluating at

x = 1

gives

u(1)=1

,

u'(1)=6

,

v(1)=4

, and

v'(1)=8

. Calculating

\frac{dy}{dx} = u'(1)v(1) + u(1)v'(1) = 6(4) + 1(8) = 32

.

Adım Adım Çözüm

1

Set up the product rule for

y = u(x)v(x)

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

and

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

The given function is a product of two composite expressions.

2

Differentiate

u(x)

using the chain rule

u'(x) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2

Differentiating the outer power function and multiplying by the derivative of the inner function

2x - 1

.

3

Differentiate

v(x)

using the chain rule

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

Differentiating the outer power function and multiplying by the derivative of the inner function

x^2 + 1

.

4

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

and evaluate at

x = 1

u(1) = 1

,

u'(1) = 6

,

v(1) = 4

,

v'(1) = 8

, giving

\frac{dy}{dx}\Big|_{x=1} = (6)(4) + (1)(8) = 32

Substituting

x = 1

into each individual term simplifies the arithmetic before combining.

Anahtar Kavram

Combined application of the Product Rule and Chain Rule

Soru 11Soru →

If $y = (2x - 3)^5$ , what is the value of $\frac{dy}{dx}$ at $x = 2$ ?

5

8

10

20

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

The derivative of

y = (2x - 3)^5

with respect to

x

requires the chain rule:

\frac{dy}{dx} = 5(2x - 3)^4 \cdot \frac{d}{dx}(2x - 3) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4

. Substituting

x = 2

gives

10(2(2) - 3)^4 = 10(1)^4 = 10

, which makes

10

the correct value.

Adım Adım Çözüm

1

Identify the inner function

u(x)

and outer function

f(u)

Let

u = 2x - 3

, so

y = u^5

.

The chain rule states that

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

.

2

Differentiate

y

with respect to

u

and

u

with respect to

x

\frac{dy}{du} = 5u^4 = 5(2x - 3)^4

and

\frac{du}{dx} = 2

.

Apply the power rule to both functions.

3

Multiply the derivatives to find

\frac{dy}{dx}

\frac{dy}{dx} = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4

.

Combine terms using the chain rule formula.

4

Substitute

x = 2

into the derivative

\frac{dy}{dx}\Big|_{x=2} = 10(2(2) - 3)^4 = 10(1)^4 = 10

.

Evaluate the expression numerically at the target point.

Anahtar Kavram

Chain Rule for Differentiation

Tahmini Süre:45s

Soru 12Soru →

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

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Cevap: 11

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is

11

.

Applying the product rule

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

along with the chain rule for the expression

(2x - 1)^4

gives

\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4

. Evaluating this at

x = 1

gives

8(1)^3(1)^3 + 3(1)^2(1)^4 = 8 + 3 = 11

.

Adım Adım Çözüm

1

Set up the product rule for

y = u \cdot v

, where

u = x^3

and

v = (2x - 1)^4

.

u = x^3

and

v = (2x - 1)^4

The function is expressed as the product of two algebraic terms.

2

Find the derivative of each function component.

\frac{du}{dx} = 3x^2

and

\frac{dv}{dx} = 8(2x - 1)^3

The power rule gives

\frac{du}{dx} = 3x^2

, and applying the chain rule to

(2x - 1)^4

yields

4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3

.

3

Substitute components into the product rule formula

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

.

\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4

Combining

u\frac{dv}{dx}

and

v\frac{du}{dx}

provides the full expression for the derivative.

4

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = 8(1)^3(2(1) - 1)^3 + 3(1)^2(2(1) - 1)^4 = 8 + 3 = 11

Substituting

x = 1

simplifies the terms to

8(1) + 3(1) = 11

.

Anahtar Kavram

Rules of Differentiation (Product and Chain Rules)

Soru 13Soru →

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 14Soru →

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 15Soru →

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 16Soru →

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 17Soru →

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 18Soru →

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

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The numerical value of the derivative at

x = 1

is 2.

To find

\frac{dy}{dx}

at

x = 1

for

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

, we use the quotient rule

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

. Setting

u = (x + 1)^3

gives

\frac{du}{dx} = 3(x + 1)^2

, and setting

v = x^2 + 1

gives

\frac{dv}{dx} = 2x

. At

x = 1

,

u = 8

,

\frac{du}{dx} = 12

,

v = 2

, and

\frac{dv}{dx} = 2

. Substituting into the quotient formula gives

\frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

.

Adım Adım Çözüm

1

Identify the components of the quotient rule

Let

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

and

v(x) = x^2 + 1

.

The given function is a quotient of two functions of

x

.

2

Differentiate the numerator using the chain rule and the denominator using standard rules

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

and

v'(x) = 2x

.

The chain rule states that

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

.

3

Evaluate all function components at

x = 1

u(1) = 8

,

u'(1) = 12

,

v(1) = 2

, and

v'(1) = 2

.

Substituting

x = 1

simplifies the calculation before applying the full quotient expression.

4

Apply the quotient rule formula to calculate the final derivative value

\frac{dy}{dx}\Big|_{x=1} = \frac{v(1)u'(1) - u(1)v'(1)}{[v(1)]^2} = \frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

.

Substituting the numerical values yields the final result.

Anahtar Kavram

Combining the Quotient Rule and Chain Rule for differentiation

Soru 19Soru →

If $y = x^2 \sqrt{3x + 1}$ , what is the value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

\frac{19}{4}

\frac{17}{4}

\frac{13}{4}

\frac{3}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{19}{4}$

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is

\frac{19}{4}

.

Applying the product rule to

y = x^2 \sqrt{3x + 1}

gives

\frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\sqrt{3x + 1}) + \sqrt{3x + 1} \cdot \frac{d}{dx}(x^2)

. By the chain rule,

\frac{d}{dx}(\sqrt{3x + 1}) = \frac{3}{2\sqrt{3x + 1}}

. Substituting

x = 1

into

\frac{dy}{dx} = \frac{3x^2}{2\sqrt{3x + 1}} + 2x\sqrt{3x + 1}

gives

\frac{3}{4} + 4 = \frac{19}{4}

. Thus, the option equal to

\frac{19}{4}

is correct.

Adım Adım Çözüm

1

Identify the main differentiation rule required and split the function into two components.

Let

u = x^2

and

v = (3x + 1)^{1/2}

, so

y = u \cdot v

.

The function is a product of

x^2

and a composite function

\sqrt{3x + 1}

.

2

Differentiate

u

with respect to

x

, and use the chain rule to differentiate

v

with respect to

x

.

\frac{du}{dx} = 2x

and

\frac{dv}{dx} = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}}

.

The chain rule requires multiplying the derivative of the outer square root function by the derivative of the inner linear function

3x + 1

.

3

Apply the product rule formula

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

.

\frac{dy}{dx} = x^2 \left( \frac{3}{2\sqrt{3x + 1}} \right) + \sqrt{3x + 1} (2x)

.

The derivative of a product

uv

is

u v' + v u'

.

4

Substitute

x = 1

into the derivative expression.

\frac{dy}{dx}\Big|_{x=1} = (1)^2 \left( \frac{3}{2\sqrt{3(1) + 1}} \right) + \sqrt{3(1) + 1} \cdot 2(1) = 1 \cdot \left( \frac{3}{4} \right) + 2 \cdot 2 = \frac{3}{4} + 4 = \frac{19}{4}

.

Evaluating at

x = 1

requires simplifying

\sqrt{4} = 2

and combining fractions.

Anahtar Kavram

Combining Product Rule and Chain Rule for Differentiation

Soru 20Soru →

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

3x^2 \sin(2x) + 2x^3 \cos(2x)

3x^2 \sin(2x) + x^3 \cos(2x)

6x^2 \cos(2x)

3x^2 \sin(2x) - 2x^3 \cos(2x)

Cevabı ve açıklamayı göster

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

Cevap

The derivative

\frac{dy}{dx}

is

3x^2 \sin(2x) + 2x^3 \cos(2x)

.

Applying the product rule

\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}

with

u = x^3

and

v = \sin(2x)

yields

x^3(2\cos(2x)) + \sin(2x)(3x^2)

, which simplifies to

3x^2 \sin(2x) + 2x^3 \cos(2x)

.

Adım Adım Çözüm

1

Identify the two component functions for the Product Rule

Let

u = x^3

and

v = \sin(2x)

, so

y = uv

.

The function is a product of an algebraic term and a trigonometric term.

2

Find the derivative of each component function

\frac{du}{dx} = 3x^2 and \frac{dv}{dx} = 2\cos(2x) using the chain rule for \sin(2x).

The derivative of \sin(2x) with respect to 2x is \cos(2x), multiplied by the derivative of 2x which is 2.

3

Apply the Product Rule formula \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

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

Combining the differentiated terms according to the product rule gives the final derivative.

Anahtar Kavram

Product Rule combined with Chain Rule for trigonometric functions

Tahmini Süre:1m 30s

Rules of Differentiation (Product, Quotient, and Chain Rules)

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