How do you find f'(x) using the limit definition given #f(x) = (3 + x)/(1 - 3x) #?

1 Answer
Douglas K.
May 13, 2017

Substitute #f(x+h)# and #f(x)# into the definition.
Make a common denominator.
Combine like terms
Cancel the h from the denominator by division.
Let #hto0#

Explanation:

#f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h" [1]"#

Given: #f(x) = (x+3)/(1-3x)" [2]"#

#f(x+h) = (x+h+3)/(1-3x-3h)" [3]"#

Substitute equations [2] and [3] into equation [1]:

#f'(x) = lim_(hrarr0) ((x+h+3)/(1-3x-3h)-(x+3)/(1-3x))/h#

We need to make a common denominator within the numerator:

#f'(x) = lim_(hrarr0) ((x+h+3)/(1-3x-3h)(1-3x)/(1-3x)-(x+3)/(1-3x)(1-3x-3h)/(1-3x-3h))/h#

Multiply the numerators within the numerator:

#f'(x) = lim_(hrarr0) ((x+h+3-3x^2-3xh-9x)/((1-3x-3h)(1-3x))-(x+3-3x^2-9x-3xh-9h)/((1-3x)(1-3x-3h)))/h#

Combine over 1 denominator:

#f'(x) = lim_(hrarr0) (((x+h+3-3x^2-3xh-9x)-(x+3-3x^2-9x-3xh-9h))/((1-3x)(1-3x-3h)))/h#

Everything sums to 0 except the h terms:

#f'(x) = lim_(hrarr0) ((10h)/((1-3x-3h)(1-3x)))/h#

Please observe the h cancels by division:

#f'(x) = lim_(hrarr0) ((10cancel(h))/((1-3x-3h)(1-3x)))/cancel(h)#

#f'(x) = lim_(hrarr0) 10/((1-3x-3h)(1-3x))#

It is safe to let #hrarr0#:

#f'(x) = 10/((1-3x)(1-3x)) = 10/(1-3x)^2#

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