find the derivative of f(x)=(3)/(x-2) at x=4
f(x)=3u^-1 where u=x-2
We will be happy to critique your work.
rewrite
f(x) = 3(x-2)^-1
then f'(x) = -3(x-2)^-2 or -3/(x-2)^2
take it from here
1. f(4+h)-f(4)/h
2. (3/(4+h)-2)-(3/2) all over h
3. then im lost
you didn't say to find the derivative by "first principles"
dy/dx = Lim[f(4+h) - f(4)]/h as h ---> 0
= lim[3/(4+h -2) - 3/2]/h as h --->0
= lim[(6 - 3h - 6)/(2(h+2)]/h as h--> 0
= lim -3h/(2(h+2))/h as h -->0
= lim -3/(2(h+2)) as h -- >0
= =3/4
To find the derivative of the function f(x) = 3/(x - 2) at x = 4, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then its derivative can be found as:
f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2
Now let's break down the steps to find the derivative of f(x) using the quotient rule:
Step 1: Identify g(x) and h(x):
In our function f(x) = 3/(x - 2), g(x) = 3 and h(x) = (x - 2).
Step 2: Find g'(x) and h'(x):
Since g(x) = 3, g'(x) = 0 (the derivative of a constant is zero).
Since h(x) = (x - 2), h'(x) = 1 (the derivative of x is 1).
Step 3: Plug the values into the quotient rule formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2
= (0 * (x - 2) - 3 * 1)/((x - 2))^2
= (-3)/(x - 2)^2
So, the derivative of f(x) = 3/(x - 2) at x = 4 is f'(4) = (-3)/(4 - 2)^2 = -3/2^2 = -3/4.