Finding Derivative of a Point?

g(x) = 4x² - 2x + 15 ← this is a function

g(5) = 100 - 10 + 15 = 105 → the representative curve of the function passes through (5 ; 105)


g'(x) = 8x - 2 ← this is its derivative

g'(5) = 40 - 2 = 38 ← this is in fact the slope of the tangent line to the curve at point (5 ; 105)


g(x) = 4x² - 2x + 15 ← this is the function → to calculate the derivative:


Lim [g(x₀ + h) - g(x₀)] / h

h → 0


Lim [ { 4.(x₀ + h)² - 2.(x₀ + h) + 15 } - { 4x₀² - 2x₀ + 15 } ] / h

h → 0


Lim [ { 4.(x₀² + 2x₀.h + h²) - 2x₀ - 2h + 15 } - 4x₀² + 2x₀ - 15 ] / h

h → 0


Lim [4x₀² + 8x₀.h + 4h² - 2x₀ - 2h + 15 - 4x₀² + 2x₀ - 15] / h

h → 0


Lim [8x₀.h + 4h² - 2h] / h

h → 0


Lim h.[8x₀ + 4h - 2] / h

h → 0


Lim (8x₀ + 4h - 2) = 8x₀ - 2

h → 0


g'(x₀) = 8x - 2

g'(5) = 40 - 2 = 38 ← this is the same result (above)

g(x) = 4x² - 2x + 15

Find the first derivative of this function, which will give you a function:

g'(x) = 8x - 2

Now substitute 5 in for x and simplify:

g'(5) = 8(5) - 2
g'(5) = 40 - 2
g'(5) = 38

If you look at the original curve at x = 5, that section will have a slope equal to 38.

Okay.  g(5) = 4(5²) - 2(5) + 15 = 100 - 10 + 15 = 105
g(5 + h) = 4(25 + 10h + h²) - 2(5 + h) + 15 = 100 + 40h + 4h² - 10 - 2h + 15
= 105 + 38h + 4h²
g(5 + h) - g(5) = 4h² + 38h
[g(5 + h) - g(5)]/h = 4h +38
The limit of (4h + 38) as h goes to zero, is 38.

g(x) = 4x^2 - 2x + 15 so, cheating temporarily
g’(x) = 8x - 2 and g’(5) = 8*5 – 2 = 38

You want use limits so here are some helpful bits you can include.
g(x + h) = 4(x + h)^2 – 2(x + h) + 15 ... (you expand that)
g(x) = 4x^2 - 2x + 15
g(x + h) – g(x) = 8hx - 2h + 4h^2
T = [g(x + h) – g(x)]/h = 8x – 2 + 4h
The limit of T as h tends to zero is 8x – 2 which is dg/dx
g’(5) = 8*5 – 2 = 38