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Question
Find the derivative of the following functions by the first principle: `1/(2x + 3)`
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Solution
Let f(x) = `1/(2x + 3)`
∴ f(x + h) = `1/(2(x + "h") + 3) = 1/(2x + 2"h"+ 3)`
By first principle, we get
f ‘(x) = `lim_("h" → 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" → 0) (1/(2x + 2"h"+ 3) - 1/(2x + 3))/"h"`
= `lim_("h" → 0) 1/"h"[(2x + 3 - 2x - 2"h" - 3)/((2x + 2"h" + 3)(2x + 3))]`
=`lim_("h" → 0) 1/"h"[(-2"h")/((2x + 2"h" + 3)(2x + 3))]`
=`lim_("h" → 0)(-2)/((2x + 2"h" + 3)(2x + 3))` …[∵ h → 0, ∴h ≠ 0]
= `(-2)/((2x + 2 xx 0 + 3)(2x + 3))`
= `(-2)/(2x + 3)^2`
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