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:
\[\lim_{x->2} \frac{x^3 + 2}{x + 1}\]
Evaluate the following:
`lim_(x->∞) (sum "n")/"n"^2`
Show that f(x) = |x| is continuous at x = 0.
Find the derivative of the following function from the first principle.
x2
If f(x)= `{((x - |x|)/x if x ≠ 0),(2 if x = 0):}` then show that `lim_(x->1)`f(x) does not exist.
Evaluate: `lim_(x->1) ((2x - 3)(sqrtx - 1))/(2x^2 + x - 3)`
For what value of x, f(x) = `(x+2)/(x-1)` is not continuous?
`"d"/"dx" (1/x)` is equal to:
If y = x and z = `1/x` then `"dy"/"dx"` =
`"d"/"dx" ("a"^x)` =
