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