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

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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.
The derivative of $$ F(x) = \left(\frac{5x - 3}{4 - 7x}\right)^6 $$ is $$ F'(x) = -6 \cdot \frac{(5x - 3)^5}{(4 - 7x)^7} $$.

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

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