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Question
Find the derivative of the following w. r. t. x by using method of first principle:
log (2x + 5)
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Solution
Let f(x) = log (2x + 5)
∴ f(x + h) = log [2(x + h) + 5]
= log (2x + 2h + 5)
By first principle, we get
f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
`lim_("h" -> 0) (log(2x + 2"h" + 5) - log(2x + 5))/"h"`
= `lim_("h" -> 0) 1/"h" log ((2x + 2"h" + 5)/(2x + 5))`
= `lim_("h" -> 0) 1/"h" log((2x + 5 + 2"h")/(2x + 5))`
= `lim_("h" -> 0) 1/"h" log (1 + (2"h")/(2x + 5))`
= `lim_("h" -> 0) log (1 + (2"h")/(2x + 5))^(1/"h")`
= `lim_("h" -> 0) log[(1 + (2"h")/(2x + 5))^((2x + 5)/(2"h"))]^(2/(2x + 5))`
= `log "e"^(2/(2x + 5)) ...[because lim_(x -> 0) (1 + "p"x)^(1/("p"x)) = "e"]`
= `2/(2x + 5) log"e"`
= `2/(2x + 5)` ...[∵ log e = 1]
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