Find the derivative of each of the following from the first principle .
(i) `sqrt(2x+3)`, (ii) `sqrt(4-x)` , (iii) `(1)/(sqrt(x))`

Explanation

(i) Let `y = sqrt(2x+3)`.
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = sqrt(2(x+deltax) + 3)`
`rArr deltay = sqrt(2(x+delta) + 3) - sqrt(2x+3)`
`rArr (deltay)/(deltax) = (sqrt(2(x+deltax) +3) - sqrt(2x+))/(deltax)`
`rArr (dy)/(dx) = underset(deltax rarr0)("lim") (deltay)/(deltax) = underset(deltaxrarr0)("lim") (sqrt(2(x+deltax) + 3) - sqrt(2x+3))/(deltax)`
`=underset(deltaxrarr0)("lim"){((sqrt(2x+2deltax + 3)- sqrt(2x+3)))/(deltax) xx ((sqrt(2x+2deltax +3) + sqrt(2x+3)))/((sqrt(2x+2deltax+3) + sqrt(2x+3)))}`
`= underset(deltaxrarr0)("lim')({(2x+2deltax+3)-(2x+3)})/(deltax.(sqrt(2x+2deltax+3) + sqrt(2x+3)))`
`= underset(deltax rarr0)("lim') (2deltax)/(deltax.(sqrt(2x+2deltax+ 3)+ sqrt(2x+3)))`
`= underset(deltax rarr0)("lim") (2)/((sqrt(2x+2deltax+3) + sqrt(2x+3)))`
`= (2)/(2sqrt(2x+3)) = (1)/(sqrt(2x+3))`.
`:. (d)/(dx) (sqrt(2x+3)) = 1/(sqrt(2x+3))`
(ii) Let `y = sqrt(4-x)`.
Let `deltax` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = sqrt(4-(x+deltax))`
`rArr (deltay)/(deltax) = sqrt(4-(x+deltax)) - sqrt(4-x)`
`rArr (deltay)/(deltax) = (sqrt(4-(x+deltax) - sqrt(4-x)))/(deltax)`
`rArr (dy)/(dx) = underset(deltax rarr0)("lim") (deltay)/(deltax)`
`= underset(deltax rarr0)("lim") ({sqrt(4-(x+deltax)) - sqrt(4-x)})/(deltax)`
` = underset(deltax rarr0)("lim") {((sqrt(4-x-deltax))-sqrt(4-x))/(deltax)xx((sqrt(4-x-deltax)+sqrt(4-x)))/((sqrt(4-x+deltax)+sqrt(4-x)))}`
`= underset(deltax rarr0) ("lim")({(4-x-deltax)- (4-x)})/(deltax(sqrt(4-x-deltax) + sqrt(4-x)))`
`= underset(deltax rarr0)("lim") (-deltax)/(deltax(sqrt(4-x-deltax) + sqrt(4-x)))`
`= underset(deltax rarr 0)("lim") (-1)/((sqrt(4-x-deltax ) + sqrt(4-x))) = (-1)/(2sqrt(4-x))`
`:. d/(dx) (sqrt(4-x)) = (-1)/(2sqrt(4-x))`
(iii) Let `y = 1/(sqrt(x))`.
Let `deltay` be an increment in y. corresponding to an increment `deltax` in x. brgt Then, `y + delta y = (1)/(sqrt(x+deltax))`
`rArr deltay = (1/(sqrt(x+delta)) - (1)/(sqrt(x)))`
`rArr (deltay)/(deltax) = (1)/(deltax) .[(1)/(sqrt(x+deltax)) - 1/(sqrt(x))]`
`rArr (dy)/(dx) = underset(deltaxrarr0)("lim") (deltay)/(deltax)`
`= underset(deltax rarr0)("lim") (1)/(deltax).[(1)/(sqrt(x+deltax))- 1/(sqrt(x))]`
`= underset(deltax rarr 0)("lim"){((sqrtx- sqrt(x+deltax)))/(deltax.sqrt(x+deltax).sqrt(x)) xx ((sqrt(x) + sqrt(x+ deltax)))/((sqrtx +sqrt(x+deltax)))}`
`= underset(deltax rarr 0)("lim") [({x-(x+deltax)})/(deltax.sqrt(x+deltax).sqrt(x).(sqrt(x)+sqrt(x+deltax)))}`
`= underset(deltax rarr 0)("lim") {(-deltax)/(sqrt(x+deltax) . sqrt(x) . (sqrt(x) + sqrt(x+deltax)))}`
`= (-1)/(sqrt(x).sqrt(x).(2sqrt(x))) = (-1)/(2^(3/2))`
Hence, `(d)/(dx) = (1/(sqrt(x))) = (-1)/(2x^(3//2))`.