We have to find the derivative of (x+3)/(2x-5).

This can be done using the product rule which states that the derivative of h(x) = f(x)* g(x) = f'(x)*g(x) + f(x)*g'(x) and the chain rule

Take the given expression (x+3)/(2x-5)

= (x+3)*(2x - 5)^-1

The derivative is (x +3)*(-1)*2*(2x - 5)^-2...

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We have to find the derivative of (x+3)/(2x-5).

This can be done using the product rule which states that the derivative of h(x) = f(x)* g(x) = f'(x)*g(x) + f(x)*g'(x) and the chain rule

Take the given expression (x+3)/(2x-5)

= (x+3)*(2x - 5)^-1

The derivative is (x +3)*(-1)*2*(2x - 5)^-2 + 1*(2x - 5)^-1

=> -2(x + 3)/ (2x - 5)^2 + 1/ (2x - 5)

=> [-2x - 6 + 2x - 5]/ (2x - 5)^2

=> -11 / (2x - 5)^2

**The required result is -11 / (2x - 5)^2**

Let f(x) = (x+3)/(2x-5)

We will use the quotient rule to find the derivative.

We will assume that f(x) = u/v such that:

u= x+3 ==> u' = 1

v = 2x-5 ==> v' = 2

==> Then, we know that f'(x) = (u'v- uv')/v^2

==> f'(x) = (1*(2x-5) - (x+3)(2)]/ (2x-5)^2

==> f'(x) = ( 2x-5 - 2x -6)/ (2x-5)^2

= ( -11/ (2x-5)^2

**==> f'(x) = -11/(2x-5)^2**