Question

Question #e7a10

Answer 1

`d/dx (2x+3)/(3x+2) = -5/(3x+2)^2`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So Let `f(x) = (2x+3)/(3x+2)` then;

`\ \ \ \ \ f(x+h) = (2(x+h)+3)/(3(x+h)+2)`
`:. f(x+h) = (2x+2h+3)/(3x+3h+2)`

And so the derivative of `y=f(x)` is given by:

`\ \ \ \ \ dy/dx = lim_(h rarr 0) ( (2x+2h+3)/(3x+3h+2) - (2x+3)/(3x+2) ) / h`
`:. dy/dx = lim_(h rarr 0) ( ((2x+2h+3)(3x+2)-(2x+3)(3x+3h+2))/((3x+3h+2)(3x+2)) ) / h`
`:. dy/dx = lim_(h rarr 0) ((6x^2+6hx+9x+4x+4h+6)-(6x^2+6hx+4x+9x+9h+6))/(h(3x+3h+2)(3x+2))`
`:. dy/dx = lim_(h rarr 0) ((6x^2+6hx+9x+4x+4h+6-6x^2-6hx-4x-9x-9h-6))/(h(3x+3h+2)(3x+2))`
`:. dy/dx = lim_(h rarr 0) (-5h)/(h(3x+3h+2)(3x+2))`
`:. dy/dx = lim_(h rarr 0) (-5)/((3x+3h+2)(3x+2))`
`:. dy/dx = (-5)/((3x+2)(3x+2))`
`:. dy/dx = (-5)/((3x+2)^2)`