Use the quotient rule to find f'(x):
f(x) = [(x+4)(x-5)]/[2x(x+3)]
For the first part...do I put them together?
Can someone please complete this.
I would expand it first to get
f(x) = (x^2 - x - 20)/(2x^2 + 3x)
now use the quotient rule
f(x) = 2x^4+x^3-43x^2-60x
To find the derivative of f(x) using the quotient rule, you will need to apply the formula:
f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2
In this case, g(x) = (x+4)(x-5) and h(x) = 2x(x+3).
First, let's find g'(x) and h'(x):
g'(x) = [(x-5) + (x+4)] = 2x-1
h'(x) = 2x+3
Now, we can substitute these values into the quotient rule formula:
f'(x) = [(2x-1) * 2x(x+3) - (x+4)(x-5) * (2x+3)] / [2x(x+3)]^2
Next, simplify the numerator:
f'(x) = [2x(x+3)(2x-1) - (x+4)(x-5)(2x+3)] / [2x(x+3)]^2
Expanding the numerator:
f'(x) = [4x^3 + 12x^2 - 2x^2 - 6x - 2x + 6 - (2x^3 + 3x^2 - 8x - 20x + 15)] / [2x(x+3)]^2
Combining like terms:
f'(x) = [2x^3 + 9x^2 - 18x + 6 - (2x^3 - 5x^2 - 28x + 15)] / [2x(x+3)]^2
Now simplify further:
f'(x) = [2x^3 + 9x^2 - 18x + 6 - 2x^3 + 5x^2 + 28x - 15] / [2x(x+3)]^2
Combine like terms again:
f'(x) = [14x^2 + 10x - 9] / [2x(x+3)]^2
Therefore, the derivative of f(x) is given by f'(x) = [14x^2 + 10x - 9] / [2x(x+3)]^2.