use the limit definition to compute the derivative

Question

use the limit definition to compute the derivatives at the given value of a.

1.g(x)= 3x−5/x−2, a=1

2.h(x)= √ 3x−4 , a=2

Answer 1

i am not sure if the square root in h(x) is over the whole binomial or just 3x, so i will just solve for the derivative of g(x).

g(x) = (3x – 5)/(x – 2)

g'(x) = lim h → 0 [f(x + h) – f(x)] / h

to keep things simple, h will become (1/h) multiplied by the remainder of the equation and remember to carry your limit all the way down.

g'(x) = 3(x + h) – 5 _ 3x – 5

----------------- --------

(x + h) – 2 x – 2

= 3x + 3h – 5 _ 3x – 5

-------------- -------- ⇒ cross multiply to find the least common denominator

x + h – 2 x – 2

= (x – 2)(3x + 3h – 5) – (3x – 5)(x + h – 2)

-------------------------------------------------------

(x + h – 2)(x – 2)

= 3x² + 3xh – 5x – 6x – 6h + 10 – (3x² + 3xh – 6x – 5x – 5h + 10)

------------------------------------------------------------------------------------ ⇒ distribute the negative and simplify

x² – 4x + xh – 2h + 4 the numerator

= 3x² + 3xh – 5x – 6x – 6h + 10 – 3x² – 3xh + 6x + 5x + 5h – 10

-----------------------------------------------------------------------------------

x² – 4x + xh – 2h + 4

= lim h → 0 (1/h) [(-h)/(x² – 4x + xh – 2h + 4) ⇒ the h in the numerator of your main fraction and the h we

put back both cancel out. evaluate your limit now.

= lim h → 0 (-1) / (x² – 4x + xh – 2h + 4) = [-1] / [x² – 4x + x(0) – 2(0) + 4] = (-1) / (x² – 4x + 4)

now that we have found the derivate g'(x), we now plug in a (or x) = 1

g(1) = [-1] / [(1)² – 4(1) + 4] = -1/1 = -1