Advertisements
Advertisements
प्रश्न
Find the derivative of the following function from the first principle.
log(x + 1)
Advertisements
उत्तर
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)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
\[\lim_{x->∞} \frac{2x + 5}{x^2 + 3x + 9}\]
Evaluate the following:
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
If `lim_(x->2) (x^n - 2^n)/(x-2) = 448`, then find the least positive integer n.
If f(x) = `(x^7 - 128)/(x^5 - 32)`, then find `lim_(x-> 2)` f(x)
Examine the following function for continuity at the indicated point.
f(x) = `{((x^2 - 9)/(x-3) "," if x ≠ 3),(6 "," if x = 3):}` at x = 3
Find the derivative of the following function from the first principle.
ex
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
Show that the function f(x) = 2x - |x| is continuous at x = 0
A function f(x) is continuous at x = a `lim_(x->"a")`f(x) is equal to:
`"d"/"dx" ("a"^x)` =
