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

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Answer 1 · 2016-11-01T00:12:58

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

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 #