Find the derivative of the function \( F(x)=\left(\frac{5 x-3}{4-7 x}\right)^{6} \)

We are given the function   F(x) = ((5x - 3)/(4 - 7x))⁶. To find F'(x), we can use the chain rule. Let   u(x) = (5x - 3)/(4 - 7x). Then F(x) = [u(x)]⁶, so by the chain rule:   F'(x) = 6[u(x)]⁵ · u'(x). Now, we need the derivative u'(x). Since u(x) is a quotient, we'll use the quotient rule. The quotient rule states that if u(x) = N(x) / D(x), then   u'(x) = [N'(x)D(x) - N(x)D'(x)] / [D(x)]². For our function, identify:   N(x) = 5x - 3  with  N'(x) = 5,   D(x) = 4 - 7x  with  D'(x) = -7. Plug these into the quotient rule:   u'(x) = [5(4 - 7x) - (5x - 3)(-7)] / (4 - 7x)². Simplify the numerator:   5(4 - 7x) = 20 - 35x,   (5x - 3)(-7) = -35x + 21. Thus:   Numerator = (20 - 35x) - (-35x + 21) = 20 - 35x + 35x - 21 = 20 - 21 = -1. So, u'(x) = -1 / (4 - 7x)². Returning to our original derivative:   F'(x) = 6[u(x)]⁵ · u'(x) = 6[(5x - 3)/(4 - 7x)]⁵ · (-1/(4 - 7x)²). This simplifies to:   F'(x) = -6 (5x - 3)⁵ / [(4 - 7x)⁵ · (4 - 7x)²]          = -6 (5x - 3)⁵ / (4 - 7x)⁷. So the derivative of F(x) is:   F'(x) = -6 (5x - 3)⁵ / (4 - 7x)⁷.