Question #e7a10

1 Answer
Steve M
Dec 7, 2016

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)

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