Question

Find the derivative of the following function by the first principle: `x sqrtx`

Answer 1

Let f(x) = `xsqrt x = x^(3/2)`

∴ f(x + h) = `(x + "h")^(3/2)`

By first principle, we get

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)][(x+h)^(3/2)+x^(3/2)])/(h[(x+h)^(3/2)+x^(3/2)])`

=`lim_("h" → 0) ((x + "h")^3 - x^3)/("h"[(x + "h")^(3/2) + x^(3/2)])`

=`lim_("h" → 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)("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]

= `(3x^2 + 3x xx0 + 0^2)/((x + 0)^(3/2) + x^(3/2))`

= `(3x^2)/(2x^(3/2)`

=`3/2x^(1/2)`

= `3/2sqrtx`