Answer 1

Let f(x) = `(x - 1)/(2x + 7)`

∴ f(x + h) = `(x + "h" - 1)/(2(x + "h") + 7) = (x + "h" - 1)/(2x + 2"h" + 7)`

By first principle, we get

f ‘(x) = `lim_("h" → 0) ("f"(x + "h") - "f"(x))/"h"`

= `lim_("h" → 0) ((x + "h" + 1)/(2x + 2"h" + 7) - (x - 1)/(2x + 7))/"h"`

= `lim_("h" → 0)1/"h" [((x + "h" - 1)(2x + 7) - (x - 1)(2x + 2"h" + 7))/((2x + 2"h" + 7) (2x + 7))]`

= `lim_("h" → 0)1/"h"[((2x^2 + 2x"h" - 2x + 7x + 7"h" - 7 - 2x^2 - 2x"h" - 7x + 2x + 2"h" + 7))/((2x + 2"h" + 7)(2x + 7))]`

= `lim_("h" → 0)1/"h"[(9"h")/((2x + 2"h" + 7)(2x + 7))]`

= `lim_("h" → 0) 9/((2x + 2"h" + 7)(2x + 7))` …[∵ h → 0, ∴ h ≠ 0]

=` 9/((2x + 2 xx 0 + 7)(2x + 7))`

= `9/(2x + 7)^2`