If f(x)=(3x+6)^–1, find f'(x).
f(u)= 1/u
f'= -1/u^2 * du
let u= 3x+6
du= 3
thank you
To find the derivative of f(x)= (3x+6)^–1, you can use the chain rule of differentiation. Here's how you can do it step by step:
Step 1: Rewrite f(x) as a negative exponent to make it easier to work with:
f(x) = (3x+6)^-1 = 1/(3x+6)
Step 2: Use the quotient rule for differentiation. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), then the derivative of f(x) is given by:
f'(x) = [h(x) * g'(x) - g(x) * h'(x)] / [h(x)]^2
For our function f(x) = 1/(3x+6), we have:
g(x) = 1 (the numerator)
h(x) = 3x+6 (the denominator)
Step 3: Find the derivatives g'(x) and h'(x):
g'(x) = 0 (since the derivative of a constant is zero)
h'(x) = 3 (the derivative of 3x+6 is 3)
Step 4: Substitute the values into the quotient rule formula:
f'(x) = [(3x+6) * 0 - 1 * 3] / [(3x+6)^2]
= -3 / (3x+6)^2
So, the derivative of f(x) is f'(x) = -3 / (3x+6)^2.