# Find the value of F'(3) when `F(x)=(f(x))/(f(x) - g(x))` and f(3) = 4, f'(3)=5, g(3)=3, g'(3)=4.

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The function F(x)=`(f(x))/(f(x) - g(x))` . f(3) = 4, f'(3)=5, g(3)=3 and g'(3)=4.

F'(x) can be determined using the quotient rule which gives:

F'(x) = `((f(x))'*(f(x) - g(x)) - f(x)*(f(x) - g(x))')/(f(x) - g(x))^2`

=> `(f'(x)*(f(x) - g(x)) - f(x)*(f'(x) - g'(x)))/(f(x) - g(x))^2`

F'(3) = `(f'(3)*(f(3) - g(3)) - f(3)*(f'(3) - g'(3)))/(f(3) - g(3))^2`

=> `(5*(4 - 3) - 4*(5 - 4))/((4 - 3))^2`

=> (5 - 4)/1

=> 1

**The value of F'(3) = 1**

F(x) = f(x) / (f(x) - g(x))

d F(x) / dx = ( (f(x) - g(x)) f'(x)-f(x) ( f'(x) - g'(x) ))/(f(x) - g(x))^2

=( f'(x)f(x) -g(x)f'(x)-f(x)f'(x)+g'(x)f(x) ) / (f(x) - g(x))^2

F'(x)=g'(x) f(x)-g(x) f'(x) / (f(x) - g(x))^2

F'(3)=g'(3) f(3)-g(3) f'(3) / (f(3) - g(3))^2

= (4x4) - (3x5) / (4-3)^2

F'(x) = (16 - 15) / 1

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