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Question
Find the derivative of the following w. r. t. x. at the point indicated against them by using method of first principle:
`2^(3x + 1)` at x = 2
Solution
Let f(x) = `2^(3x + 1)`
∴ f(2) = `2^(3(2) + 1)` = 27 and
f(2 + h) = `2^(3(2 + "h") + 1) = 2^(3"h" + 7)`
By first principle, we get
f'(a) = `lim_("h" -> 0) ("f"("a" + "h") - "f"("a"))/"h"`
∴ f'(2) = `lim_("h" -> 0) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) (2^(3"h" + 7) - 2^7)/"h"`
= `lim_("h" -> 0) (2^(3"h") * 2^7 - 2^7)/"h"`
= `lim_("h" -> 0) (2^7 (2^(3"h") - 1))/"h"`
= `2^7 lim_("h" -> 0) ((2^(3"h") - 1)/(3"h")) xx 3`
= `2^7 (log 2) xx 3 ...[lim_(x -> 0) (("a"^("p"x) - 1)/("p"x)) = log "a"]`
= 384 log 2
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