Question

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, yielding f'(x) = 17 / (3x+4)^2. To find f'(1), we substitute x = 1 into the derivative, obtaining f'(1) = 17/49.

Step-by-step explanation:

The question appears to be asking how to find the derivative of a function f(x) = (2x-3)/(3x+4).

To find f'(x), which is the derivative of f(x), we use the quotient rule of differentiation.

The quotient rule states that if you have a function that is the quotient of two functions u(x)/v(x), then its derivative is given by (v(x)u'(x) - u(x)v'(x))/(v(x))^2.

For f(x) = (2x-3)/(3x+4),

u(x) = 2x-3 and

v(x) = 3x+4.

We compute u'(x) and v'(x) to be 2 and 3, respectively. Therefore:

f'(x) = [(3x+4)(2) - (2x-3)(3)] / (3x+4)^2.

After simplifying, we find:

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

= 17 / (3x+4)^2.

To find f'(1), we substitute x = 1 into f'(x):

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

= 17 / 49.