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Question
Find the derivative of the following function by the first principle: 3x2 + 4
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Solution
Let f(x) = 3x2 + 4
∴ f(x + h) = 3(x + h)2 + 4
= 3(x2 + 2xh + h2) + 4
= 3x2 + 6xh + 3h2 + 4
By first principle, we get
f ′(x) =`lim_("h"→ 0) ("f"(x + "h") - "f"(x))/"h"`
=`lim_("h" → 0) ((3x^2 + 6x"h" + 3"h"^2 + 4) - (3x^2 + 4))/"h"`
=`lim_("h" → 0) (3"h"^2 + 6x"h")/"h"`
=`lim_("h"→0) (h(3h+6x))/h`
=`lim_("h" → 0)(6x + 3"h")` …[∵ h → 0, ∴h ≠ 0]
= 6x + 3(0)
= 6x
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