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Question
Find the derivative of the following w. r. t. x by using method of first principle:
x2 + 3x – 1
Solution
Let f(x) = x2 + 3x – 1
∴ f(x + h) = (x + h)2 + 3(x + h) – 1
= x2 + 2xh + h2 + 3x + 3h – 1
By first principle, we get
f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) ([x^2 + 2x"h" + "h"^2 + 3x + 3"h" - 1] - [x^2 + 3x - 1])/"h"`
= `lim_("h" -> 0) (2x"h" + "h"^2 + 3"h")/"h"`
= `lim_("h" -> 0) ("h"(2x + "h" + 3))/"h"`
= `lim_("h" -> 0) (2x + "h" + 3)` ...[∵ h → 0, ∴ h ≠ 0]
= 2x + 3
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