Question

Given that F(x)= lim∆x_0 f(x ∆×) f(×)/∆× and f(×) = 2×-3/3× 4. Find f'(×) and f'(1)

Answer 1

The derivative of the function f(x) = (2x - 3) / (3x + 4) is found using the quotient rule, which results in f'(x) = 17 / (3x + 4)^2, and specifically, f'(1) = 17 / 49.

Step-by-step explanation:

The question posed seems to be from a misunderstanding of notation and calculus terminology. It looks like the intention is to find the derivative of the function f(x) = (2x - 3) / (3x + 4). The derivative function is usually denoted as f'(x), and finding it involves using the quotient rule in calculus, which states that if you have a function that is the quotient of two functions u(x)/v(x), the derivative f'(x) is given by:

(v(x)u'(x) - u(x)v'(x)) / (v(x))^2

Let's apply this rule to our function:

Let u(x) = 2x - 3, then

u'(x) = 2.

Let v(x) = 3x + 4, then

v'(x) = 3.

So the derivative f'(x) is:

((3x+4)(2) - (2x-3)(3)) / (3x+4)^2

Then we simplify it:

(6x + 8 - 6x + 9) / (3x+4)^2 = 17 / (3x+4)^2

To find f'(1), replace x with 1 in the derivative:

f'(1) = 17 / (3(1)+4)^2

= 17 / 49.