InterviewSolution
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InterviewSolution
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Find the derivative of the following from the first principle. (i) sin x^(2) , (ii) cos (x^(2)+ 1) (iii) tan x^(2) |
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Answer» Solution :(i)Let `y = sin^(2)X` . Let `deltay` be an increment in y, corresponding to an increment `deltax`in `x`. Then, `y + deltay = sin(x+deltax)^(2)` `rArr deltay = sin(x+deltax)^(2)` `rArr (deltay)/(deltax) =(sin(x+deltax)^(2) - sinx^(2))/(deltax)` `rArr (dy)/(dx) = underset(deltaxrarr0)("lim") (deltay)/(deltax)` `= underset(deltaxrarr0)("lim") (sin(x+deltax)^(2) - sinx^(2))/(deltax)` `= underset(deltaxrarr0)("lim") (2cos[((x+deltax)^(2) + x^(2))/(2)] sin [((x+deltax)^(2) -x^(2))/(2)])/(deltax)` [Using `[using(sin C - sinD) = 2 COS((C+D)/(2)) sin((C-D)/(2))]` `= underset(deltaxrarr0)("lim") 2cos[((x+deltax)^(2) + x^(2))/(2)] (sin[(x+(deltax)/(2)).deltax])/((x+(deltax)/(2)).deltax).(x+(deltax)/(2))` `2.underset(deltararr0)("lim")cos[((x+deltax)^(2)+x^(2))/(2)].underset(deltararr0)("lim") (sin[(x+(deltax)/(2)).deltax])/((x+(deltax)/(2)).deltax).underset(deltaxrarr0)("lim") (x+(deltax)/(2))` `= [2xx COSX^(2) xx 1 xx x] = 2x cos x^(2)`. Hence, `d/(dx) (sinx^(2)) = 2x cos x^(2)`. (ii) Let `y = cos(x^(2) + 1)` . Let `deltay` be an increment in y, corresponding to an increment`deltax` in x. Then, `y + deltay = cos[(x+deltax)^(2) + 1]` `rArr deltay = cos[(x+deltax)^(2) + 1] - cos(x^(2) + 1)` `rArr (deltay)/(deltax) = (cos[(x+deltax)^(2) + 1] - cos(x^(2) + 1))/(deltax)` `rArr (dy)/(dx) = underset(deltaxrarr0)("lim") (deltay)/(deltax)` `=underset(deltaxrarr0)("lim")({cos[(x+deltax)^(2)+1]-cos(x^(2)+1)})/(deltax)` `=underset(deltaxrarr0)("lim") (-2sin'[({(x+deltax)^(2) + 1+ (x^(2)+1)})/(2)].sin[({(x+deltax)^(2)+1-(x^(2)+1)})/(2)])/(deltax)` `=-2.underset(deltaxrarr0)("lim")sin[x^(2)+x.deltax+1+((deltax)^(2))/(2)].(sin[(x+(deltax)/(2)).deltax])/([(x+(deltax)/(2)).deltax]).(x+(deltax)/(2))` `=-2.underset(deltaxrarr0)("lim")sin[x^(2)+x.deltax+1+((deltax)^(2))/(2)].underset(thetararr0)("lim") (sintheta)/(theta).underset(deltaxrarr0)("lim") (x+(deltax)/(2))`. where `theta = (x+(deltax)/(2)).deltax` clearly, `[deltaxrarr0] rArr[theta rarr 0]` `=- 2sin(x^(2)+ 1) xx 1 xx x` `= -2x sin (x^(2) + 1)`. Hence, `d/(dx) [cos(x^(2) + 1)] = - 2xsin (x^(2) + 1)`. (iii) Let `y = TAN^(2) x^(2)` Let `deltay` be an increment in y, correponding to an increment `deltax`in x. Then, `y + deltay = tan(x+deltax)^(2)` `rArr deltaxy = tan(x+deltax)^(2) - tanx^(2)` `rArr (deltay)/(deltax) = (tan(x+deltax)^(2) - tanx^(2))/(deltax)` ` rArr (dy)/(dx) = underset(deltax rarr0)("lim") (deltay)/(deltax) = underset(deltaxrarr0)("lim") (tan(x+deltax)^(2) - tanx^(2))/(deltax)` `= underset(deltaxrarr0)("lim") ({(sin(x+deltax)^(2))/(cos(x+deltax)^(2))- (sinx^(2))/(cosx^(2))})/(deltax)` `= underset(deltaxrarr0)("lim")([sin(x+deltax)^(2)cos x^(2)-cos(x+deltax)^(2)sinx^(2)])/(deltax.cos(x+deltax)^(2).cosx^(2))` `= underset(deltaxrarr0)("lim")(sin[(x+deltax)^(2)-x^(2)])/(deltax.cos(x+deltax)^(2).cosx^(2))` `= underset(deltaxrarr0)("lim")(sin[(2x+deltax).deltax])/([(2x+deltax).deltax]) .((2x+deltax))/(cos(x+deltax)^(2).cosx^(2))` `= (underset(thetararr0)("lim") (sintheta)/(theta)). underset(deltaxrarr0)("lim") (2x+deltax). underset(deltaxrarr0)("lim") (1)/(cos(x+deltax)^(2).cosx^(2))` where `(2x+deltax). deltax = theta` [clearly, `(deltaxrarr0) rArr (theta rarr0)`] `= 1 xx 2xxx (1)/(cosx^(2).cosx^(2)) = 2x SEC^(2)x^(2)`. Hence `d/(dx) (tan x^(2)) = 2 x sec^(2) x^(2)`. |
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