solve [(2\3(x+h)-4)-(2/3x-4)]/h Find the derivative of f'(x) where x=2, and g(x)=2/3x-4

Question

I was given, for the following function, find the derivative by definition and write the equation of the line tangent to the given function at the point where x=2.

Definition of a derivative of f(x): f'(x)=lim, h=0, [(f)x+h)-f(x)]/h

 

g(x)=2/3x-4

 

I understand how substituting in works(I think), what I don't know is how to do the math to follow through with the rest of the equation..

Answer 1

g(x) = 2 / (3x - 4)

g'(x)
= lim(h->0) [[g(x + h) - g(x)] / h]
= lim(h->0) [[(2 / (3(x + h) - 4)) - (2 / (3x - 4))] / h]
= lim(h->0) [[2(3x - 4) - 2(3(x + h) - 4)] / [h(3x - 4)(3(x + h) - 4]]
= lim(h->0) [6x - 8 - 6x - 6h + 8] / [h(3x - 4)(3(x + h) - 4)]
= lim(h->0) [-6h] / [h(3x - 4)(3(x + h) - 4)]
= lim(h->0) -6 / [(3x - 4)(3(x + h) - 4)]
= -6 / [(3x - 4)(3x - 4)]
= -6 / (3x - 4)^2

Thus, slope at x = 2 will be
= g'(2)
= -6 / (3(2) - 4)^2
= -6 / (6 - 4)^2
= -6 / 2^2
= -6 / 4
= -3/2

Also, when x = 2, we will have:

y
= g(2)
= 2 / (3(2) - 4)
= 2 / 2
= 1

Using the slope formula, we have:

(y - 1) / (x - 2) = -3/2
(y - 1) = (-3/2)(x - 2)
y - 1 = (-3/2)x + 3
y = (-3/2)x + 4, which is the equation of the tangent line.