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Question
Find the derivative of the following function from the first principle.
log(x + 1)
Solution
Let f(x) = log(x + 1)
Then f(x + h) = log(x + h + 1) = log((x + 1) + h)
Now `"d"/"dx"`f(x)
`= lim_(h->0) ("f"(x + "h") - "f"(x))/"h"`
`= lim_(h->0) (log (x + 1) + "h" - log (x + 1))/"h"`
`= lim_(h->0) (log (((x + 1) + "h")/(x + 1)))/"h"`
`= lim_(h->0) (log (1 + "h"/(x + 1)))/"h"`
`= lim_(h->0) (log (1 + "h"/(x + 1)))/((("h")/(x+1)) xx (x + 1))`
`= 1/(x+1) lim_(h->0) (log (1 + "h"/(x + 1)))/("h"/(x+1))`
`"d"/"dx" "f"(x) = 1/(x + 1)`
∴ `"d"/"dx" log (x + 1) = 1/(x + 1)`
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