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Question
Find the derivative of the following w. r. t. x by using method of first principle:
`x sqrt(x)`
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Solution
Let f(x) = `x sqrt(x) = x^(3/2)`.
Then f(x + h) = `(x + "h")^(3/2)`
By definition,
f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) ((x + "h")^(3/2) - x^(3/2))/"h"`
= `lim_("h" -> 0) ((x + "h")^(3/2) - x^(3/2))/"h" * ((x + "h")^(3/2) + x^3/2)/((x + "h")^(3/2) + x^(3/2))`
= `lim_("h" -> 0) ((x + "h")^3 - x^3)/("h"[(x + "h")^(3/2) + x^(3/2)]`
= `lim_(x -> 0) (x^3 + 3x^2"h" + 3x"h"^2 + "h"^3 - x^3)/("h"[(x + "h")^(3/2) + x^(3/2)]`
= `lim_("h" -> 0) ("h"(3x^2 + 3x"h" + "h"^2))/("h"[(x + "h")^(3/2) + x^(3/2)]`
= `lim_("h" -> 0) (3x^2 + 3x"h" + "h"^2)/((x + "h")^(3/2) + x^(3/2))` ...[h → 0, h ≠ 0]
= `(lim_("h" -> 0) (3x^2 + 3x"h" + "h"^2))/(lim_("h" -> 0) (x + "h")^(3/2) + lim_("h" -> 0) x^(3/2))`
= `(3x^2 + 3x xx 0 + 0^2)/((x + 0)^(3/2) + x^(3/2))`
= `(3x^2)/(2x^(3/2)`
= `3/2 sqrt(x)`
Alternative Method:
Let f(x) = `xsqrt(x) = x^(3/2)`.
Then f(x + h) = `(x + "h")^(3/2)`
∴ f'(x) = `lim_("h" -> 0) ("f"(x + "h") - "f"(x))/"h"`
= `lim_("h" -> 0) ((x + "h")^(3/2) - x^(3/2))/"h"`
Put x + h = y
∴ h = y – x and as h → 0, y → x
∴ f'(x) = `lim_(y -> x) (y^(3/2) - x^(3/2))/(y - x)`
= `3/2 x^(3/2 - 1) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `3/2 sqrt(x)`
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