How do you use the limit definition to find the derivative of f(x)=(4-3x)/(2+x)?

1 Answer
Nov 1, 2016

f(x) = -10/(x+2)^2

Explanation:

By definition f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h )

So, with f(x)=(4-3x)/(2+x) we have:
f(x) = lim_(hrarr0)( ( ((4-3(x+h))/(2+(x+h))) - ((4-3x)/(2+x)) ) / h )

:. f(x) = lim_(hrarr0)( ( ((4-3x-3h)/(2+x+h)) - ((4-3x)/(2+x)) ) / h )

:. f(x) = lim_(hrarr0)( (( (4-3x-3h)(2+x) - (4-3x)(2+x+h) )/((2+x+h)(2+x))) / h )

:. f(x) = lim_(hrarr0)( (8+4x-6x-3x^2-6h-3hx) - (8+4x+4h-6x-3x^2-3xh) )/(h(2+x+h)(2+x))

:. f(x) = lim_(hrarr0)( 8+4x-6x-3x^2-6h-3hx - 8-4x-4h+6x+3x^2+3xh )/(h(2+x+h)(2+x))

:. f(x) = lim_(hrarr0)( ( -10h )/(h(2+x+h)(2+x)) )

:. f(x) = lim_(hrarr0)( ( -10 )/((2+x+h)(2+x)) )

:. f(x) = ( -10 )/((2+x)(2+x))

Hence, f(x) = -10/(x+2)^2