Please try to solve it?

Question

Differentiate: Product rule

1. Y= X^3 ( 2X + 3 )

2. Y= X^2 ( X + 1 )^3

3. Y= ( X^3 + 5X^2 - 3 )( X^2 + 1 )^5

4. Y= ( 5X + 3 )( X - 7)^-1/3

5. Y= ( 2X + 9 ) √(X^2 - 4)

Answer 1

Hi there...

Well our product rule states that  d/dx ( u.v ) = ( vu' + uv' ) so:

1) If we let u = x³ and v = 2x+3 then

u' = 3x² and v' = 2

d/dx(x³(2x+3)) = d/dx(u.v) = uv' + vu' = x³.2 + (2x+3).3x² = 2x³ + 6x³ + 9x² = 8x³ + 9x²

2) letting u = x² and v = (x+1)³ then

u' = 2x and v' = 3(x+1)²...now substitute into product rule:

dy/dx = vu' + uv' = (x+1)³.2x + x².3(x+1)² = 2x(x+1)³ + 3x²(x+1)²

3) let u = (x³ + 5x² - 3) and v = (x² + 1)^5

so u' = 3x² + 10x and v' = 5(x² + 1)^4 .(2x)

now substitute into your product rule for u, v, u' and v'.

4) let u=(5x + 3) and v = (x-7)^-1/3

so u' = 5 and v' = -1/3(x-7)^-4/3

now substitute into product rule.

5) let u = (2x + 9) and v = √(x² - 4) = (x² - 4)^1/2

so u' = 2 and v' = 1/2(x² - 4)^-1/2 .(2x)

now subst. into product rule.

Note: If you're unsure of how I got v' in questions 3 and 5 you should look up the chain rule for differentiating functions of functions.

Hope it helps (^_^)

Answer 2

You actually have a combination of the product rule and the chain rule here

Product rule says if y = uv then y ' = u ' v + v ' u

1. y = x³(2x + 3)

Here u = x³ so u ' = 3x²

v = 2x + 3 so v ' = 2

y ' = u ' v + v ' u

y ' = 3x²(2x + 3) + 2x³

2. y= x²(x + 1)³

Here u = x² so u ' = 2x

v = (x + 1)³ so v ' = 3(x + 1)² (chain rule)

y ' = u ' v + v ' u

y ' = 2x(x + 1)³ + 3x²(x + 1)²

3. y = (x³ + 5x² - 3)(x² + 1)^5

Here u = x³ + 5x² - 3 so u ' = 3x² + 10x

v = (x² + 1)^5 so

v ' = 5*2x*(x² + 1)⁴  (chain rule)

= 10x(x² + 1)⁴

y ' = u ' v + v ' u

y ' = (3x² + 10x)(x² + 1 )^5 + 10x(x³ + 5x² - 3)(x^2 + 1)⁴

4. y = (5x + 3)(x - 7)^(-1/3)

so u = 5x + 3 and u ' = 5

v = (x - 7)^(-1/3) and v ' = -1/3 * (x - 7)^(-4/3)

y ' = u ' v + v ' u

y ' = 5(x - 7)^(-1/3) - 1/3 * (5x + 3)(x - 7)^(-4/3)

5. y = (2x + 9)√(x² - 4)

u = 2x + 9

u ' = 2

v = √(x² - 4) = (x² - 4)^½

v ' = 1/2 * 2x * (x² - 4)^(-½) ... chain rule

= x / √(x² - 4)

y ' = u ' v + v ' u

y ' = 2√(x² - 4) + x(2x + 9) / √(x² - 4)