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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Find `(dy)/(dx)`, when: `x^(n)+y^(n)=a^(n)` |
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Answer» Correct Answer - `(-x^(n-1))/(y^(n-1))` `x^(n)+y^(n)=a^(n)rArrnx^(n-1)+ny^(n-1)(dy)/(dx)=0.` |
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| 152. |
Find `(dy)/(dx)`, when: `cot(xy)+xy=y` |
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Answer» Correct Answer - `(-ycot^(2)(xy))/({1+xcot^(2)(xy)})` `cot(xy)+xy=y` `rArr-"cosec"^(2)(xy).{x(dy)/(dx)+y}+(x(dy)/(dx)+y)=(dy)/(dx)` `rArr{"cosec"^(2)(xy)-1)}.x(dy)/(dx)+(dy)/(dx)={1-"cosec"^(2)(xy)}y` `rArr{xcot^(2)(xy)+1}.(dy)/(dx)=-ycot^(2)(xy).` |
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| 153. |
If `ln((e-1)e^(xy) +x^2)=x^2+y^2` then `((dy)/(dx))_(1,0)` is equal to |
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Answer» `"We have, "(e-1)e^(xy)+x^(2)=e^(x^(2))+y^(2)` Differentiating both sides w.r.t. x, we get `(e-1)cdote^(xy)cdot(xcdot(dy)/(dx)+y)+2x=e^(x^(2)+y^(2)).(2x+2y(dy)/(dx))` Putting x=1 and y = 0, we get `(e-1)((dy)/(dx))+2=e(2+0)` `rArr" "((dy)/(dx))_((1,0))=2` |
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| 154. |
if `y=((a-x)sqrt(a-x)-(b-x)sqrt(x-b))/(sqrt(a-x)+sqrt(x-b))` then `dy/dx` wherever it is defined is equal to`:`A. `(x+(a+b))/(sqrt((a-x)(x-b)))`B. `(2x-a-b)/(2sqrt(a-x)sqrt(x-b))`C. `-((a+b))/(2sqrt((a-x)(x-b)))`D. `(2x+(a+b))/(2sqrt((a-x)(x-b)))` |
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Answer» `y=((a-x)^(3//2)+(x-b)^(3//2))/(sqrt(a-x)+sqrt(x-b))` `=((sqrt(a-x)+sqrt(x-b))(a-x-sqrt(a-x)sqrt(x-b)+x-b))/(sqrt(a-x)+sqrt(x-b))` `=a-b-sqrt(a-x)sqrt(x-b)` `therefore" "(dy)/(dx)=(1)/(2sqrt(a-x))sqrt(x-b)-(1)/(2sqrt(x-b))sqrt(a-x)` `=(2x-a-b)/(2sqrt(a-x)sqrt(x-b))` |
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| 155. |
the derivative of `y=(1-x)(2-x)..........(n-x)` at `x=1` is equal to |
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Answer» `(dy)/(dx)=-[(2-x)(3-x)...(n-x)+(1-x)(3-x)...(n-x)+...(1-x)(2-x)...(n-1-x)]` At x = 1, `(dy)/(dx)=-[(n-1)!+0+...+0]=(-1)(n-1)!` |
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| 156. |
`d/(dx)sqrt((1-sin2x)/(1+sin2x))i se q u a lto,(0A. `sec^(2)x`B. `-sec^(2)((pi)/(4)-x)`C. `sec^(2)((pi)/(4)+x)`D. `sec^(2)((pi)/(4)-x)` |
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Answer» `y=sqrt((1-sin 2x)/(1+ sin 2x))=(cos x - sin x)/(cos x + sin x)=(1-tan x)/(1+ tan x)=tan ((pi)/(4)-x)` `therefore (dy)/(dx)=-sec^(2)((pi)/(4)-x)` |
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| 157. |
`(d^2x)/(d y^2)` equalsA. `-((d^(2)y)/(dx^(2)))((dy)/(dx))^(-3)`B. `((d^(2)y)/(dx^(2)))^(-1)`C. `-((d^(2)y)/(dx^(2)))^(-1)((dy)/(dx))^(-3)`D. `((d^(2)y)/(dx^(2)))((dy)/(dx))^(-2)` |
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Answer» We have `(d^(2)x)/(dy^(2))=(d)/(dy)=((dx)/(dy))=(d)/(dy)(((1)/(dy))/(dx))cdot(dx)/(dy)` `=-(1)/(((dy)/(dx))^(2))cdot(d^(2)y)/(dx^(2))cdot(1)/(((dy)/(dx)))` `=-(1)/(((dy)/(dx))^(3))cdot(d^(2)y)/(dx^(2))` `=-((dy)/(dx))^(-3)((d^(2)y)/(dx^(2)))` |
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| 158. |
`(1)/((x^(2)-3x+5)^(3))` |
| Answer» Correct Answer - `(-3(2x-3))/((x^(2)-3x+5)^(4))` | |
| 159. |
`sqrt((a^(2)-x^(2))/(a^(2)+x^(2)))` |
| Answer» Correct Answer - `(-2a^(2)x)/((a^(2)+x^(2))^(3//2)(a^(2)-x^(2))^(1//2))` | |
| 160. |
If `y=sin^(2)alpha+cos^(2)(alpha+beta)+2sinalphasinbetacos(alpha+beta)`, then `(d^(3)y)/(dalpha^(3))`, isA. `(sin^(3)(alpha+beta))/(cosalpha)`B. `cos(alpha+3beta)`C. 0D. none of these |
| Answer» Correct Answer - C | |
| 161. |
If `f(x) = |cos x| + |sin x|`, then `dy/dx` at `x = (2pi)/3` is equal toA. `(1-sqrt(3))/(2)`B. 0C. `(1)/(2)(sqrt(3)-1)`D. none of these |
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Answer» In neighborhood of `x=(2pi)/(3),| cos x| = -cos x and | sin x| = som x` `"or "y=-cos x+ sin x` `therefore" "(dy)/(dx)=sin x + cos x` Thus, at `x=(2pi)/(3)," we get "(dy)/(dx)= sin""(2pi)/(3)+cos""(2pi)/(3)=(sqrt(3)-1)/(2).` |
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| 162. |
Find `(dy)/(dx)" for "y=sin^(-1) (cos x), x in (0, pi)cup (pi, 2pi).` |
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Answer» We have, `sin^(-1)(cos x)=(pi)/(2)-cos^(-1)(cos x)` `={:{((pi)/(2)-"x,"," "if 0ltxltpi),((pi)/(2)-(2pi-x)","," "if piltxlt2pi):}` `={:{((pi)/(2)-"x,"," "if 0ltxltpi),(x-(3pi)/(2)-(2pi-x)","," "if piltxlt2pi):}` `therefore" "(d)/(dx){sin^(-1)(cos x)}={:{(-1","," "if 0ltxltpi),(1","," "if piltxlt2pi):}` |
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| 163. |
Let `z=(cosx)hat5a n dy="sin"xdot`Then the value of `2(d^2z)/(dy^2)a tx=(2pi)/9`is____________. |
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Answer» `z=(cos x)^(5),y=sin x` `(dz)/(dx)=-5cos^(4)xcdot sin x, (dy)/(dx)= cos x` `therefore" "(dz)/(dy)=-5cos^(3)x cdot sin x` `"Now, "(d^(2)z)/(dy^(2))=(d)/(dx)((dz)/(dy))cdot(dx)/(dy)` `=-5(d)/(dx)[cos^(3)xcdot sin x] (1)/(cos x)` `=-5[cos^(4)x- 3 sin^(2)xcdot cos^(2)x ](1)/(cos x)` `=-5(cos^(3)x-3 sin^(2)xcdotcos x)` `=-5(cos^(3)x-3 cos x(1-cos^(2)x))` `=-5(4 cos^(3)x-3 cos x)` `=-5 cos 3x` `therefore" "(d^(2)z)/(dy^(2)):|_(x=(2pi)/(9))=-5 cos 120^(@)=(5)/(2)` |
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| 164. |
`cos^(2)x^(3)` |
| Answer» Correct Answer - `-3x^(2)sin(2x^(3))` | |
| 165. |
`"If "y^(x)=x^(y)," then find "(dy)/(dx).` |
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Answer» Correct Answer - `(y(x log y-y))/(x(y log x -x))` `y^(x)=x^(y)` `"or "log y^(x)= log x^(y)` `"or "x log y = y log x` `"or "x(1)/(y)(dy)/(dx)+log y xx1 =(y)/(x)+log x (dy)/(dx)` `"or "((x)/(y)-log x)(dy)/(dx)=(y)/(x)-log y` `"or "(dy)/(dx)=(y/x-logy)/((x)/(y)-log x)=(y(y-x log y))/(x(x- y log x))=(y(x log y - y ))/(x(y log x -x))` |
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| 166. |
If sec (x+y) = xy, then find `(dy)/(dx)` |
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Answer» We have, sec (x+y)=xy On differentiating both sides w.r.t x, we get `sec (x+y)cdot tan (x+y) cdot (d)/(dx) (x+y) = x (dy)/(dx)+y` `rArr overset(cdot)sec (x+y) cdot tan (x+y) cdot (1+(dy)/(dx))=x(dy)/(dx)+y` `rArr (dy)/(dx)[sec (x+y)cdot tan (x+y)-x]` `=y-sec (x+y)cdot. tan (x+y)` `therefore (dy)/(dx)=(y-sec(x+y).tan (x+y))/(sec (x+y). tan (x+y)-x` |
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| 167. |
Find `(dy)/(dx)`, when: `y=((x+1)^(2)sqrt(x-1))/((x+4)^(3).e^(x))` |
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Answer» Correct Answer - `((x+1)^(2)sqrt(x-1))/((x+4)^(3).e^(x)).{(2)/((x+1))+(1)/(2(x-1))-(3)/((x+4))-1}` `logy=2log(x+1)+(1)/(2)log(x-1)-3log(x+4)-x` `rArr(1)/(y).(dy)/(dx)={(2)/((x+1))+(1)/(2(x-1))-(3)/((x+4))-1}.` |
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| 168. |
`sec^(3)(x^(2)+1)` |
| Answer» Correct Answer - `6x sec^(3)(x^(2)+1)tan(x^(2)+1)` | |
| 169. |
`sqrt(cos3x)` |
| Answer» Correct Answer - `(-3)/(2).(sin3x)/(sqrt(cos 3x))` | |
| 170. |
Find `(dy)/(dx)`, when: `y=(x^(5)sqrt(x+4))/((2x+3)^(2))` |
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Answer» Correct Answer - `(x^(5)sqrt(x+4))/((2x+3)^(2)).{(5)/(x)+(1)/(2(x+4))-(4)/((2x+3))}` `logy=5logx+(1)/(2)log(x+4)-2log(2x+3)` `rArr(1)/(y).(dy)/(dx)={(5)/(x)+(1)/(2(x+4))-(4)/((2x+3))}.` |
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| 171. |
Find `(dy)/(dx)`, when: `y=cosxcos2x cos3x` |
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Answer» Correct Answer - `-(cosx cos2x cos3x).[tanx+2 tan 2x+3 tan 3x]` `logy=log cosx+logcos2x+log cos3x` `rArr(1)/(y).(dy)/(dx)=(-sinx)/(cosx)-(2sin2x)/(cos2x)-(3sin3x)/(cos3x)` `rArr(dy)/(dx)=y.{-tanx-2 tan2x-3 tan 3x}.` |
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| 172. |
`y=sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5))),f i n d(dy)/(dx)` |
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Answer» Correct Answer - `(1)/(2)sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5))).[(1)/((x-1))+(1)/((x-2))-(1)/(x-3)-(1)/((x-4))-(1)/((x-5))]` `logy=(1)/(2).{log(x-1)+log(x-2)-log(x-3)-logx(-4)-log(x-5)}` `rArr(1)/(y).(dy)/(dx)=(1)/(2).{(1)/((x-1))+(1)/((x-2))-(1)/((x-3))-(1)/((x-4))-(1)/((x-5))}.` |
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| 173. |
Find `(dy)/(dx)`, when: `y=sin(x^(x))` |
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Answer» Correct Answer - `[cos(x^(x))]x^(x)(1+logx)` Let `x^(x)=t`. Then, `logt=xlogx rArr(1)/(t).(dt)/(dx)=x.(1)/(x)+logx.1` `rArr(dt)/(dx)=t[1+logx]=x^(x)(1+logx).` `thereforey=sint rArr(dy)/(dt)=cost=cosx^(x).` `therefore(dy)/(dx)=((dy//dt))/((dx//dt))=[cos(x^(x))]x^(x)(1+logx).` |
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| 174. |
`sec(x+y) = xy` |
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Answer» We have, sec (x+y)=xy On differentiating both sides w.r.t x, we get `sec (x+y)cdot tan (x+y) cdot (d)/(dx) (x+y) = x (dy)/(dx)+y` `rArr overset(cdot)sec (x+y) cdot tan (x+y) cdot (1+(dy)/(dx))=x(dy)/(dx)+y` `rArr (dy)/(dx)[sec (x+y)cdot tan (x+y)-x]` `=y-sec (x+y)cdot. tan (x+y)` `therefore (dy)/(dx)=(y-sec(x+y).tan (x+y))/(sec (x+y). tan (x+y)-x` |
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| 175. |
If `y=1=x/(1!)+(x^2)/(2!)+(x^3)/(3!)++(x^n)/(n !),`show that `(dy)/(dx)-y+(x^n)/(n !)=0.` |
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Answer» `(dy)/(dx)=0+(1)/(1!)+(1)/(2!)(2x)+(1)/(3!)(3x^(2))+...+(1)/(n!)(nx^(n-1))` `=1+(x)/(1!)+(x^(2))/(2!)+...+(x^(n-1))/((n-1)!)` `={1+(x)/(1!)+(x^(2))/(2!)+...+(x^(n-1))/((n-1)!)+(x^n)/(n!)}-(x^(n))/(n!)` `=y-(x^(n))/(n!)` `"or "(dy)/(dx)-y+(x^(n))/(n!)=0` |
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| 176. |
Find `(dy)/(dx)`for `y=sin^(-1)(cosx),`where `x in (0,2pi)dot` |
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Answer» We have, `sin^(-1)(cos x)=(pi)/(2)-cos^(-1)(cos x)` `={:{((pi)/(2)-"x,"," "if 0ltxltpi),((pi)/(2)-(2pi-x)","," "if piltxlt2pi):}` `={:{((pi)/(2)-"x,"," "if 0ltxltpi),(x-(3pi)/(2)-(2pi-x)","," "if piltxlt2pi):}` `therefore" "(d)/(dx){sin^(-1)(cos x)}={:{(-1","," "if 0ltxltpi),(1","," "if piltxlt2pi):}` |
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| 177. |
If `cos^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=tan^(-1)a`, prove than `(dy)/(dx)=(y)/(x).` |
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Answer» `(x^(2)-y^(2))/(x^(2)+y^(2))=cos [tan^(-1)a]=" constant"` `rArr(d)/(dx)((x^(2)-y^(2))/(x^(2)+y^(2)))=0` `rArr((x^(2)+y^(2)).(d)/(dx)(x^(2)-y^(2))-(x^(2)-y^(2)).(d)/(dx)(x^(2)+y^(2)))/((x^(2)+y^(2))^(2))=0` `rArr(x^(2)+y^(2))[2x-2y(dy)/(dx)]-(x^(2)-y^(2))(2x+2y(dy)/(dx))=0` `rArrx{(x^(2)+y^(2))-(x^(2)-y^(2))}=y{(x^(2)=y^(2))+(x^(2)+y^(2))}(dy)/(dx).` Hence, `(dy)/(dx)=(y)/(x).` |
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| 178. |
If `y=logsqrt((1+sin^(2)x)/(1-sinx))`, find `(dy)/(dx)`. |
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Answer» We have `y=(1)/(2){log(1+sin^(2)x)-log(1-sinx)}." ...(i)"` On differentiating (i) w.r.t. x, we get `(dy)/(dx)=(1)/(2).[(d)/(dx){log(1+sin^(2)x)}-(d)/(dx){log(1-sinx)}]` `=(1)/(2).{(2sinx cos x)/(1+sin^(2)x)-((-cos x))/((1-sinx))}` `=(1)/(2).{(sin2x)/((1+sin^(2)x))+(cosx)/((1-sinx))}`. |
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| 179. |
If `y=(xsin^(-1)x)/(sqrt(1-x^2))+logsqrt(1-x^2)`, then prove that `(dy)/(dx)=(sin^(-1)x)/((1-x^2)^(3/2))` |
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Answer» Let `x = sin theta => dx/(d theta) = cos theta ` Now, `y = (theta sin theta)/sqrt(1-sin^2theta) + log sqrt(1-sin^2theta)` `=>y = (theta sin theta)/cos theta +log cos theta` `=>y = theta tan theta +log cos theta` `=>dy/(d theta) = tan theta + theta(sec^2 theta)+1/costheta(-sintheta)` `=>dy/(d theta) = tan theta + theta/(cos^2theta)-tan theta` `=>dy/(d theta) = theta/(cos^2theta)` `=> dy/dx = (dy/(d theta))/(dx/(d theta)) = (theta/(cos^2theta))/(cos theta)` `=> dy/dx = (theta/(cos^3 theta))` Now, `x = sin theta => cos theta = sqrt(1-x^2)` `=> dy/dx = (sin^-1x)/(1-x^2)^(3/2).` |
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| 180. |
`y=logsqrt(sinsqrt(e^(x)))` |
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Answer» Correct Answer - `(1)/(4)e^(x//2)cot (e^(x//2))` `(d)/(dx)[logsqrt(sin)sqrt(e^(x))]=(d)/(dx)[(1)/(2)log (sinsqrt(e^(x)))]` `=(1)/(2)cot sqrt(e^(x))(1)/(2sqrt(e^(x)))e^(x)=(1)/(4)e^(x//2)cot (e^(x//2))` |
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| 181. |
`"If "sqrt(x)+sqrt(y)=4," then find "(dy)/(dx).` |
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Answer» We have `sqrt(x)+sqrt(y)=4` Differentiating the given equation both sides w.r.t.x, we get `(1)/(2sqrt(x))+(d)/(dx)(sqrt(y))=0` `rArr(1)/(2sqrt(x))+((d)/(dy)(sqrt(y)))(dy)/(dx)=0` `rArr(1)/(2sqrt(x))+(1)/(2sqrt(y))cdot(dy)/(dx)=0` `rArr(dy)/(dx)=-(sqrt(y))/(sqrt(x))` |
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| 182. |
If `y=x+1/(x+1/(x+1/(x+ dot)))`, prove that `(dy)/(dx)=y/(2y-x)`. |
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Answer» We have `y=x + (1)/(x + (1)/(x + (1)/(x+....)))=x+(1)/(y)` `or y^(2)=xy+1` ` or 2y(dy)/(dx)=y+x(dy)/(dx)+0" [Differentiating both sides w.r.t.x]"` `or (dy)/(dx)(2y-x)=y` `or (dy)/(dx)=(y)/(2y-x)` |
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| 183. |
Find `(dy)/(dx)`, when `y=e^(a x)cos(b x+c)` |
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Answer» We have `(dy)/(dx)=e^(ax).(d)/(dx){cos(bx+c)}+cos(bx+c).(d)/(dx)(e^(ax))` `=e^(ax).{-b sin(bx+c)}+cos(bx+c).ae^(ax)` `=e^(ax).{a cos (bx+c)-b sin (bx+c)}`. |
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| 184. |
Differentiate `logsqrt((1+cos^(2)x)/((1-e^(2x))))` w.r.t. x. |
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Answer» Let `y=logsqrt((1+cos^(2)x)/((1-e^(2x))))=log((1+cos^(2)x)/(1-e^(2x)))^(1//2)=(1)/(2)log((1+cos^(2)x)/(1-e^(2x)))` `thereforey=(1)/(2)log(1+cos^(2)x)-(1)/(2)log(1-e^(2x))` `rArr(dy)/(dx)=(1)/(2).(1)/((1+cos^(2)x))(2 cos x)(-sinx)-(1)/(2).(1)/((1-e^(2x))).(-2e^(2x))` `={(-sinx cos x)/((1+cos^(2)x))+(e^(2x))/((1-e^(2x)))}`. Hence, `(d)/(dx){logsqrt((1+cos^(2)x)/((1-e^(2x))))}={(-sin x cos x)/((1+cos^(2)x))}+(e^(2x))/((1-e^(2x)))}.` |
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| 185. |
If `y=sec(tan^(-1)x)`, then `(tan^(-1)x)`, then `(dy)/(dx)" at "x=1` is equal toA. `(1)/(sqrt(2))`B. `-(1)/(sqrt(2))`C. 1D. none of these |
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Answer» Correct Answer - A We have, `y=sec(tan^(-1)x)` `implies" "y=sec(sec^(-1)sqrt(1+x^(2))` `implies" "y=sqrt(1+x^(2))implies(dy)/(dx)=(x)/(sqrt(1+x^(2)))implies((dy)/(dx))_(x=1)=(1)/(sqrt(2))` |
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| 186. |
The function `u=e^x s in ; v=e^x cos x`satisfy the equation`v(d u)/(dx)-u(d v)/(dx)=u^2+v^2`b. `(d^2u)/(dx^2)=2v`c. `(d^2)/(dx^2)=-2u`d. `(d u)/(dx)+(d v)/(dx)=2v`A. `v(du)/(dx)u(dv)/(dx)=u^(2)+v^(2)`B. `(d^(2)u)/(dx^(2))=2v`C. `(d^(2)v)/(dx^(2))=-2u`D. all the above |
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Answer» Correct Answer - D We have, `u=e^(x)sinximplies(du)/(dx)=e^(x)sinx+e^(x)cosx=u+v` `v=e^(x)cosximplies(dv)/(dx)=e^(x)cosx+e^(x)sinx=v-u` `:." "v(du)/(dx)-(udv)/(dx)=v(u+v)-u(v-u)=u^(2)+v^(2)` `(d^(2)u)/(dx^(2))=(du)/(dx)+(dv)/(dx)=u+v+v-u=2v` and,`(d^(2)u)/(dx^(2))=(dv)/(dx)-(du)/(dx)=(v-u)-(v+u)=-2u` |
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| 187. |
`"Let "y=t^(10)+1 and x=t^(8)+1." Then "(d^(2)y)/(dx^(2))` isA. `(5)/(2)t`B. `20t^(8)`C. `(5)/(16t^(6))`D. none of these |
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Answer» `"Here, "y=t^(10)+1 and x=t^(8) +1` `therefore" "t^(8)=x-1 or t^(2)=(x-1)^(1//4)` `"So, "y=(x-1)^(5//4)+1` Differentiating both sides w.r.t. x, we get `(dy)/(dx)=(5)/(4)(x-1)^(1//4)` Again, differentiating both sides w.r.t. x, we get `(d^(2)y)/(dx^(2))=(5)/(16)(x-1)^(-3//4)=(5)/(16(x-1)^(3//4))=(5)/(16(t^(2))^(3))=(5)/(16t^(6))` |
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| 188. |
If `f(x)=x^4tan(x^3)-x1n(1+x^2),`then the value of`(d^4(f(x)))/(dx^4)`at `x=0`is0 (b) 6(c) 12 (d)24 |
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Answer» `"As "f(x) = x^(4) tan (x^(3))-x" In "(1+x^(2))" is odd, "(d^(3)f(x))/(dx^(3))` is even `"Hence, "(d^(4)f(x))/(dx^(4))=0at x = 0.` |
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| 189. |
If x = log p and `y=(1)/(p),` thenA. `(d^(2)y)/(dx^(2))-2p=0`B. `(d^(2)y)/(dx^(2))+y=0`C. `(d^(2)y)/(dx^(2))+(dy)/(dx)=0`D. `(d^(2)y)/(dx^(2))-(dy)/(dx)=0` |
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Answer» `(dy)/(dx)=(-(1)/(p^(2)))/((1)/p)=-(1)/(p)=-y` `"or "(d^(2)y)/(dx^(2))=-(dy)/(dx)` `"or "(d^(2)y)/(dx^(2))+(dy)/(dx)=0` |
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| 190. |
If `y=(x+sqrt(x^2+a^2))^n ,t h e n(dy)/(dx)`is`(n y)/(sqrt(x^2+a^2))`(b) `-(n y)/(sqrt(x^2+a^2))``(n x)/(sqrt(x^2+a^2))`(d) `-(n x)/(sqrt(x^2+a^2))`A. `(ny)/(sqrt(x^(2)+a^(2))`B. `-(ny)/(sqrt(x^(2)+a^(2))`C. `(nx)/(sqrt(x^(2)+a^(2))`D. `-(nx)/(sqrt(x^(2)+a^(2))` |
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Answer» `(dy)/(dx)=(d)/(dx)[(x+sqrt(x^(2)+a^(2)))^(n)]` `=n(x+sqrt(x^(2)+a^(2)))^(n-1).(d)/(dx)(x+sqrt(x^(2)+a^(2)))` `n=(x+sqrt(x^(2)+a^(2)))^(n-1)((sqrt(x^(2)+a^(2))+x)/(sqrt(x^(2))+a^(2)))` `=(n(x+sqrt(x^(2)+a^(2)))^(n))/(sqrt(x^(2))+a^(2))` `=(ny)/(sqrt(x^(2))+a^(2))` |
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| 191. |
If ( sin x) (cos y) = `1//2,` then `d^(2)y//dx^(2)` at `(pi//4, pi//4)` isA. `-4`B. `-2`C. `-6`D. 0 |
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Answer» `(sin x ) (cos y )=(1)/(2)` `therefore" "(cos x ) (cos y) - sin y sin x (dy)/(dx) =0` `"or "(dy)/(dx)=(cot x )(cot y)` `"or "(d^(2)y)/(dx^(2))=-cosec^(2)xcdotcot y -cosec^(2)y cot xcdot (dy)/(dx)` `"Now, "((dy)/(dx))_(((pi//4,pi//4)))=1` `"or "((d^(2)y)/(dx^(2)))_(((pi//4,pi//4)))=-(2) (1) -(2) (1) (1) =-4` |
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| 192. |
If `x=a (theta-sin theta) and y=a(1-cos theta)`, find `(d^(2)y)/(dx^(2))` at `theta=pi.` |
| Answer» Correct Answer - `(-1)/(4a)` | |
| 193. |
`(d)/(dx)cos^(-1)sqrt(cosx),0ltxlt(pi)/(2)` is equal toA. `(1)/(2)sqrt(1+sec x)`B. `sqrt(1+sec x)`C. `-(1)/(2)sqrt(1+sec x)`D. `-sqrt(1+sec x)` |
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Answer» `(d)/(dx)cos^(-1)sqrt(cos x)=(sin x)/(2sqrt(cos x)sqrt(1-cos x))` `=(sqrt(1- cos^(2)x))/(2sqrt(cos x)sqrt(1-cos x))=(1)/(2)sqrt((1+cos x)/(cos x))` |
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| 194. |
If xy = 1, prove that \(\cfrac{dy}{d\mathrm x}\) + y2 = 0 |
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Answer» We are given with an equation x y = 1, we have to prove that \(\cfrac{dy}{d\mathrm x}\)+ y2 = 0 by using the given equation we will first find the value of \(\cfrac{dy}{d\mathrm x}\)and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to x, we get, By using product rule, we get, y(1) + x\(\cfrac{dy}{d\mathrm x}\) = 0 \(\cfrac{dy}{d\mathrm x}\) = \(\cfrac{-y}{\mathrm x}\) Or we can further solve it by using the given equation, \(\cfrac{dy}{d\mathrm x}\) = \(\cfrac{-y}{\frac{1}y}\) = -y2 By putting this value in the L.H.S. of the equation, we get, –y2 + y2 = 0 = R.H.S. |
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| 195. |
Find the second - order derivative of : `(i)x^(10)" "(ii)logx" "(iii)tan^(-1)x` |
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Answer» (i) Let `y=x^(10)`. Then, `(dy)/(dx)=10x^(9).` `therefore (d^(2)y)/(dx^(2))=(10xx9)x^(8)=90x^(8)`. Hence, `(d^(2)y)/(dx^(2))(x^(10))=90^(8).` (ii) Let `y=logx.` Then, `(dy)/(dx)=(1)/(x)=x^(-1)`. `therefore (d^(2)y)/(dx^(2))=(d)/(dx)(x^(-1))=(-1)x^((-1)-1)=-x^(-2)=(-1)/(x^(2)).` Hence, `(d^(2))/(dx^(2))(logx)=(-1)/(x^(2)).` (iii) Let `y=tan^(-)x.` Then, `(dy)/(dx)=(1)/((1+x^(2)))=(1+x^(2))^(-1)`. `therefore(d^(2)y)/(dx^(2))=(d)/(dx)(1+x^(2))^(-1)=(-1)(1+x^(2))^(-2).(2x)=(-2x)/((1+x^(2))^(2))`. Hence, `(d^(2))/(dx^(2)){tan^(-1)x}=(-2x)/((1+x^(2))^(2))`. |
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| 196. |
If `y=sin(logx)`, prove that `x^2(d^2y)/(dx^2)+x(dy)/(dx)+y=0`. |
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Answer» We have `y=sin(logx)` `rArr(dy)/(dx)=(d)/(dx){sin(logx)}=cos(logx).(1)/(x)=(cos(logx))/(x)` `rArr(d^(2)y)/(dx^(2))=(d)/(dx){(cos(logx))/(x)}` `=(x.(d)/(dx){cos(logx)}-cos(logx).(d)/(dx)(x))/(x^(2))` `=(x{-sin(logx).(1)/(x)}-cos(logx).1)/(x^(2))` `=(-{sin(logx)+cos(logx)})/(x^(2)).` |
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| 197. |
If `t=(x^(4)+cotx),` find `(d^(2)y)/(dx^(2))`. |
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Answer» We have `y=(x^(4)+cotx)` `rArr (dy)/(dx)=4x^(3)-"cosec"^(2)x` `rArr (d^(2)y)/(dx^(2))=(d)/(dx)(4x^(3)-"cosec"^(2)x)` `=4.(d)/(dx)(x^(3))-(d)/(dx)("cosec"^(2)x)` `=(4xx3x^(2))-2"cosec x"(-"cosec x cot x")` `=(12x^(2)+2"cosec"^(2)xcot x).` |
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| 198. |
If`y=1/(1+x^(a-b)+x^(c-b))+1/(1+x^(b-c)+x^(a-c))+1/(1+x^(b-a)+x^(c-a)),`then (dy)/(dx) is equal to(a)`1`(b) `(a+b+c)^(x+b+c-1)``0`(d) none of these |
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Answer» `y = 1/(1 + x^a/x^b + x^c/x^b) + 1/(1 + x^b/x^c + x^a/x^c) + 1/(1+x^b/x^a + x^c/x^a)` `y = x^b/(x^a + x^b + x6c) + x^c/(x^a + x^b + x^c) + x^a/(x^a + x^b + x^c) ` `= (x^b + x^c + x^a )/(x^a + x^b + x^c) = 1` `y = 1` `dy/dx = 0` option c is correct |
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| 199. |
Find second orderderivative of `log(logx)` |
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Answer» Let `y=log(logx).` Then, `(dy)/(dx)=(1)/((logx)).(1)/(x)=(1)/((xlogx))=(xlogx)^(-1)` `rArr(d^(2)y)/(dx^(2))=(d)/(dx)(x logx)^(-1)` `=(-1)(xlogx)^(-2).(d)/(dx)(xlogx)` `=(-1)/((xlogx)^(2)).(x.(1)/(x)+logx.1)=(-(1+logx))/((xlogx)^(2)).` |
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| 200. |
Differentiate `tan^(-1){(5x)/(1-6 x^2)}, -1/(sqrt(6)) |
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Answer» Correct Answer - `((3)/(1+9x^(2))+(1)/((1+x^(2))))` `y=tan^(-1)3x+tan^(-1)2x.` |
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