If f(x)=[(6x^3)-3]/x^4
a)find f' (x)
b) find f'(4)
To find the derivative of the function f(x)=[(6x^3)-3]/x^4, you can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are both differentiable functions, the derivative of f(x) is given by:
f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2
Let's break down the problem step by step:
a) To find f'(x), we need to find the derivatives of g(x) and h(x) and substitute them into the quotient rule formula.
g(x) = 6x^3 - 3
g'(x) = 18x^2 (Take the derivative of 6x^3, which is 18x^2, and the derivative of -3, which is 0)
h(x) = x^4
h'(x) = 4x^3 (Take the derivative of x^4, which is 4x^3)
Now, plug these values into the quotient rule formula:
f'(x) = [(18x^2 * x^4) - (6x^3 - 3)*(4x^3)] / [(x^4)]^2
Simplify the expression:
f'(x) = [18x^6 - (24x^6 - 12x^3)] / x^8
f'(x) = [18x^6 - 24x^6 + 12x^3] / x^8
f'(x) = [-6x^6 + 12x^3] / x^8
f'(x) = -6x^6/x^8 + 12x^3/x^8
f'(x) = -6/x^2 + 12/x^5
Therefore, f'(x) = -6/x^2 + 12/x^5.
b) To find f'(4), we substitute x = 4 into the expression we found for f'(x):
f'(4) = -6/(4)^2 + 12/(4)^5
f'(4) = -6/16 + 12/1024
f'(4) = -3/8 + 3/256
f'(4) = -768/2048 + 24/2048
f'(4) = -744/2048
Therefore, f'(4) is equal to -744/2048.