Problem : Calculate the derivative of f (x) = x ^{2} at x = 1 .
Substituting 1 for x _{0} in the formula for the derivative, we have
f'(1) | = | ||
= | |||
= | 2 + Δx | ||
= | 2 |
Problem : Find the vertex of the parabola f (x) = x ^{2} + 2x + 2 using the derivative.
At the vertex, the tangent line to the graph will be horizontal, with slope 0 . Therefore, we search for an x such that f'(x) = 0 . We have
f'(x) | = | limΔx→0 | |
= | limΔx→0 | ||
= | limΔx→0 | ||
= | limΔx→02x + 2 + Δx | ||
= | 2(x + 1) |
Problem : Find the equation of the tangent line to the graph of f (x) = x ^{3} at x = 2 .
First we compute f'(2) :
f'(2) | = | limΔx→0 | |
= | limΔx→0 | ||
= | 12 |