Question

Help. Please its urgent show workings.​

Answer 1

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

$\frac{dy}{dx}$ = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × $\frac{d}{dx}$ (x + 4) ← chain rule

       = 2(x + 4) × 1

        = 2(x + 4)

Then

$\frac{dy}{dx}$ = (2x - 1). 2(x + 4) + (x + 4)². 2

    = 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

    = 2(x + 4) (2x - 1 + x + 4)

    = 2(x + 4)(3x + 3) ← factor out 3

    = 6(x + 4)(x + 1)

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(b)

y =  x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × $\frac{d}{dx}$ (x² - 1) ← chain rule

        = 3(x² - 1)² × 2x

        = 6x(x² - 1)²

Then

$\frac{dy}{dx}$ = x. 6x(x² - 1)² + (x² - 1)³. 1

    = 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

    = (x² - 1)² (6x² + x² - 1)

     = (x² - 1)²(7x² - 1)

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(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

$\frac{dy}{dx}$ = (x² - 1). 3x² + (x³ + 1), 2x

   = 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

   = x[3x(x² - 1) + 2(x³ + 1) ]

   = x(3x³ - 3x + 2x³ + 2)

   = x(5x³ - 3x + 2) ← distribute

    = 5$x^{4}$ - 3x² + 2x

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(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × $\frac{d}{dx}$(x² + 4) ← chain rule

       = 2(x² + 4) × 2x

       = 4x(x² + 4)

Then

$\frac{dy}{dx}$ = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

    = 12$x^{4}$(x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)

    = 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

    = 3x²(x² + 4)(4x² + 3x² + 12)

    = 3x²(x² + 4)(7x² + 12)