InterviewSolution
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InterviewSolution
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Differentiate each of the following from the first principle . (i) sin sqrt(x) , (ii) cos sqrt(x) (iii) tan sqrt(x) |
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Answer» Solution :(i) Let `y = sin sqrt(X)` Let `deltay` be an increment in y, corresponding to an increment `deltax` in x. Then, `y +deltay=sinsqrt(x+deltax)` `rArr deltay=sinsqrt(x+deltax)-sinsqrt(x)` `rArr (deltay)/(deltax)=(sinsqrt(x+deltax)-sinsqrt(x))/(deltax)` `rArr(dy)/(dx) = underset(deltaxrarr0)("lim")(deltay)/(deltax)` `= underset(deltaxrarr0)("lim")({sinsqrt(x+deltax) - sinsqrt(x)})/(deltax)` `= underset(deltararr0)("lim") (2cos((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/({(x+deltax) - x})` `["USING" sinX - sinD = 2 cos' ((C+D))/(2) sin ((C-D)/(2))]` `= underset(deltaxrarr0)("lim") (2cos((sqrt(x+deltax) + sqrt(x))/(2))sin ((sqrt(x+deltax) - sqrt(x))/(2)))/((sqrt(x+deltax) + sqrt(x)) (sqrt(x+deltax) - sqrt(x)))` `[ :'(x+deltax) -x = (sqrt(x+deltax) + sqrt(x))(sqrt(x+deltax)-sqrt(x))]` `= underset(deltaxrarr0)("lim")(sin((sqrt(x+deltax) - sqrt(x))/(2)))/(((sqrt(x+deltax)-sqrt(x))/(2))). underset(deltaxrarr0)("lim")cos((sqrt(x+deltax)+sqrt(x))/(2)). underset(deltaxrarr0)("lim") (1)/((sqrt(x+deltax) + sqrt(x)))` `= underset(thetararr0)("lim")(sintheta)/(theta).underset(deltaxrarr0)("lim")(cos(sqrt(x+deltax) + sqrt(x))/(2)).underset(deltaxrarr0)("lim")(1)/((sqrt(x+deltax)+sqrt(x)))`, where `((sqrt(x+deltax) - sqrt(x)))/(2) = theta` and `deltax rarr 0 rArr theta rarr 0` `= 1 xx cos sqrt(x) xx 1/(2sqrt(x))= (cossqrt(x))/(2sqrt(x))`. HENCE , `d/(dx) (sinsqrt(x)) = (cossqrt(x))/(2sqrt(x))`. (i) Let `y = cos sqrt(x)`. Let `deltay` be an increment in y, corresponding to an increment `deltax`in x. Then, `y+deltay = cossqrt(x + deltax) - cos sqrt(x)` `rArr(deltay)/(deltax) = ({cossqrt(x+deltax)-cossqrt(x)})/(deltax)` `rArr (dy)/(dx) = underset(deltaxrarr0)("lim")(deltay)/(deltax)` `= underset(deltax rarr 0)("lim") ({cossqrt(x+deltax)- cos(x)})/(deltax)` ` = underset(deltaxrarr0)("lim") (-2sin((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/({(x+deltax) - x})` ` = underset(deltaxrarr0)("lim") (-2sin((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/((sqrt(x+deltax)+sqrt(x))(sqrt(x+deltax)-sqrt(x)))` `=-underset(deltaxrarr0)("lim")sin((sqrt(x+deltax)+sqrt(x))/(2)).underset(deltararr0)("lim") (sin '((sqrt(x+ deltax) - sqrt(x))/(2)))/(((sqrt(x+deltax)-sqrt(x))/(2))) . underset(deltararr0)("lim")(1)/((sqrt(x+deltax)+sqrt(x)))` where `((sqrt(x+deltax)-sqrt(x)))/(2) = theta` `= - sin((2sqrt(x))/(2)) xx 1 xx 1/(2sqrt(x) ) = (-sinsqrt(x))/(2sqrt(x))` Hence , d/(dx) (cossqrt(x)) = (-sinsqrt(x))/(2sqrt(x))`. (iii) Let `y = TAN sqrt(x)`. Let `deltay` be an increment in y, corresponding to an increment `deltax` in x. Then, `y + deltay = tan sqrt(x+deltax)` `rArr deltay = tan sqrt(x+deltax) - tan sqrt(x)` `rArr (deltay)/(deltax) = (tan(sqrt(x+deltax)- tan sqrt(x)))/(deltax)` `rArr (dy)/(dx) = underset(deltax rarr 0)("lim") (deltay)/(deltax)` `= underset(deltax rarr 0)("lim") ((tan sqrt(x+deltax) - tan sqrt(x)))/(deltax)` `= underset(deltaxrarr0)("lim")({(sinsqrt(x+deltax))/(cos sqrt(x+deltax))-(sinsqrt(x))/(cossqrt(x))})/(deltax)` `= underset(deltaxrarr0)("lim") ((sinsqrt(x+deltax) cos sqrt(x) - cos sqrt(x+deltax) sinsqrt(x)))/(deltax.cos sqrt(x+deltax).cossqrt(x))` ` = underset(deltaxrarr0)("lim") {(sin(sqrt(x+deltax) - sqrt(x)))/(|(x+deltax)-x|) . (1)/(cossqrt(x+deltax).cossqrt(x))}` `[ :' deltax = (x+deltax )- x]` `= underset(deltaxrarr0)("lim") {(sin(sqrt(x+deltax)-sqrt(x)))/((sqrt(x+deltax) - sqrt(x))(sqrt(x+deltax) + sqrt(x))).(1)/(cos sqrt(x+deltax).cossqrt(x))}` `=underset(deltararr0)("lim")(sin(sqrt(x+deltax) - sqrt(x)))/((sqrt(x+deltax) - sqrt(x))).underset(deltaxrarr0)("lim") (1)/((sqrt(x+deltax)+sqrt(x))).underset(deltaxrarr0)("lim") (1)/(cossqrt(x+deltax).cossqrt(x))` `= (underset(thetararr0)("lim") (sintheta)/(theta)).1/(2sqrt(x)).(1)/(cos^(2)sqrt(x))=1xx 1/(2sqrt(x))xx(1)/(cos^(2)sqrt(x))` `= (sec^(2)sqrt(x))/(2sqrt(x))`. Hence, `d/(dx) (tansqrt(x)) = (sec^(2)sqrt(x))/(2sqrt(x))`, |
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