Differentiate the function. y = (4x^2 + 6x + 2)/√x

Answer 1

Hi Emanuel,

You can follow rule of fraction f(x) / g(x) =[f'(x) g(x) - f(x) g'(x)] ⁄ g^2(x)

y = (4x^2 + 6x + 2)/√x f(x) = 4x^2 + 6x + 2 and g(x) = √x

for better understanding first find derivative

1) f'(x) = 8x +6

2) g'(x) = 1/2√x

now replace to the formula

y' = [ ( 8x +6 ) √x - (4x^2 +6x +2)1/2√x] / (√x)^2

y' = (8x√x + 6√x - 2x^1.5 -3x^.5 - 1/√x ) / x some times you need to simplify

now, every terms must be divided by x

y' = 8x√x /x + 6√x /x - 2x^1.5/x -3x^.5/x - 1/x√x

y' = 8√x + 6/√x -2√x -3/√x - 1/x√x

y' = 6√x + 3/√x - 1/x√x

I hope it is useful,

Minoo

Answer 2

For this you have two options. Treat it like as either a product rule (4x^2+6x+2) * x^(-1/2) OR treat it as a quotient rule. I always use the rhyme to remember the order (low d-high minus high d-low, square the bottom and away you go), as it doesn't rhyme if you have the wrong order.

For this I'll be treating the function as a product rule as there's no order to remember.

let h(x) = f(x) * g(x), then h'(x) = f'(x)g(x) + g'(x)f(x)

let f(x) = 4x^2 + 6x +2 and g(x) = x^(-1/2)

f'(x) = 8x + 6 and g'(x) = (-1/2)x^(-3/2)

Put those together:

h'(x) = (8x+6)(x^(-1/2)) + [(-1/2)x^(-3/2)*(4x^2 + 6x + 2)]

and simplifying a little gets you

h'(x) = 6x^(1/2) + 3x^(-1/2) - x^(-3/2)

When using the product and quotient rules, it can be helpful to write out every step and tackle the problem in smaller bite-sized pieces even when the process becomes second nature. To this day, I explicitly write out each function and it's derivative to keep my work organized and neat and maximize partial credit if I make a mistake.