Find f'(x): f(x)=(x^2-3x)/x^2
A. 2x-3/x^2
B. 2x-3/2x
C. 1-(3/x)
D. 3/x^2
E. None of these
I can put it into the division rule for the derivative but get confused when I'm simplifying.
To find the derivative of the function f(x) = (x^2 - 3x) / x^2, you can use the quotient rule. The quotient rule states that if you have a function in the form f(x) = g(x) / h(x), where g(x) and h(x) are differentiable functions, then the derivative f'(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply the quotient rule to find f'(x) for the given function.
First, we need to find g(x) and h(x). In this case, g(x) = x^2 - 3x and h(x) = x^2.
Next, calculate the derivatives of g(x) and h(x) separately:
g'(x) = 2x - 3 (derivative of x^2 is 2x, and derivative of -3x is -3)
h'(x) = 2x (derivative of x^2 is 2x)
Now substitute the values into the quotient rule formula:
f'(x) = ((2x - 3) * x^2 - (x^2 - 3x) * 2x) / (x^2)^2
= (2x^3 - 3x^2 - 2x^3 + 6x^2) / x^4
= (3x^2 + 6x^2) / x^4
= 9x^2 / x^4
= 9 / x^2
So, the correct answer is D. f'(x) = 9 / x^2.