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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.
| 1. |
`"If "y^(1//m)=(x+sqrt(1+x^(2)))," then "(1+x^(2))y_(2)+xy_(1)` is (where `y_(r)` represents the rth derivative of y w.r.t. x)A. `m^(2)y`B. `my^(2)`C. `m^(2)y^(2)`D. none of these |
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Answer» We have `y^(1//m)=(x+sqrt(1+x^(2)))` `"or "y=(x+sqrt(1+x^(2)))^(m)` `"or "(dy)/(dx)=m(x+sqrt(1+x^(2)))^(m-1)(1+(x)/(sqrt(x^(2)+1)))` `=m((x+sqrt(1+x^(2)))^(m))/(sqrt(1+x^(2)))` `=(my)/(sqrt(1+x^(2)))` `"or "y_(1)^(2)(1+x^(2))=m^(2)y^(2)` `"or "2y_(1)y_(2)(1+x^(2))+2xy_(1)^(2)=2m^(2)yy_(1)` `"or "y_(2)(1+x^(2))+xy_(1)=m^(2)y` |
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| 2. |
Find `(dy)/(dx)`, when: `y=e^(x)sin^(3)xcos^(4)x` |
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Answer» Correct Answer - `e^(x)sin^(3)xcos^(4)x(1+3cot x-4 tanx)` `logy=x+3 log sin x+4log cos x` `rArr(1)/(y).(dy)/(dx)={1+(3cosx)/(sinx)-(4)/(cosx).sinx}.` |
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| 3. |
If y= sin px and `y_(n)` is the nth derivative of y, then `|{:(y,y_(1),y_(2)),(y_(3),y_(4),y_(5)),(y_(6),y_(7),y_(8)):}|`isA. 1B. 0C. -1D. none of these |
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Answer» `D=|{:( sin px, p cos px, -p^(2) sin px),(-p^(3)cos px, p^(4) sin px, p^(5) cos px),(-p^(6) sin px ,-p^(7)cos px , p^(8) sin px):}|` `=p^(9)|{:(sin px, p cos px, -p^(2) sin px),(-cos px ,p sin px , p^(2) cos px),(-sin px , -p cos px, -p^(2)sin px):}|` `=-p^(9)|{:(sin px , p cos px, -p^(2) sin px),(cos px , p sin px , p^(2) cos px),(sin px, p cos px,-p^(2)sin px):}|=0` |
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| 4. |
Find `(dy)/(dx)`, when: `y=2^(x).e^(3x)sin4x` |
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Answer» Correct Answer - `2^(x).e^(3x)sin4x.{(log2)+3+4cot4x}` `logy=xlog2+3x+log sin 4x` `rArr(1)/(y).(dy)/(dx)=(log2)+3+(4cos4x)/(sin4x).` |
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| 5. |
Find the second derivative of `e^(3x)sin4x.` |
| Answer» Correct Answer - `e^(3x)(24cos4x-7sin 4x)` | |
| 6. |
If for `x (0,1/4),`the derivative of `tan^(-1)((6xsqrt(x))/(1-9x^3))`is `sqrt(x)dotg(x),`then `g(x)`equals:`(3x)/(1-9x^3)`(2) `3/(1+9x^3)`(3) `9/(1+9x^3)`(4) `(3xsqrt(x))/(1-9x^3)`A. `(9)/(1+9x^(3))`B. `(3xsqrt(x))/(1-9x^(3))`C. `(3x)/(1-9x^(3))`D. `(3)/(1+9x^(3))` |
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Answer» Correct Answer - A Let `y=tan^(-1)((6xsqrt(x))/(1-9x^(3))),x in(0,(1)/(4))`. Then, `y=tan^(-1)((2xx3xsqrt(x))/(1-(3xsqrt(x))^(2)))` `implies" "y=2tan^(-1)(3xsqrt(x))" "[{:(becausetan^(-1)((2x)/(1-x^(2)))=2tan^(-1)x","),(" if "-1ltxlt1):}]` `implies" "(dy)/(dx)=2xx(1)/(1+9x^(3))xx(9)/(2)sqrt(x)=sqrt(x)(9)/(1+9x^(3))` `implies" "(dy)/(dx)=sqrt(x)g(x)," where "g(x)=(1)/(1+9x^(3))` |
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| 7. |
The slope of the tangent to the curve `(y-x^5)^2=x(1+x^2)^2`at the point `(1,3)`is. |
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Answer» `(y-x^(5))^(2)=x(1+x^(2))^(2)` Differentiating both sides w.r.t. x, we get `2(y-x^(5))((dy)/(dx)-5x^(4))=1(1+x^(2))^(2)+(x)(2(1+x^(2))(2x))` On putting x=1 , y =3 in above equation, we get `(dy)/(dx)=8` |
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| 8. |
If for `x (0,1/4),`the derivative of `tan^(-1)((6xsqrt(x))/(1-9x^3))`is `sqrt(x)dotg(x),`then `g(x)`equals:`(3x)/(1-9x^3)`(2) `3/(1+9x^3)`(3) `9/(1+9x^3)`(4) `(3xsqrt(x))/(1-9x^3)`A. `(3)/(1+9x^(3))`B. `(9)/(1+9x^(3))`C. `(3xsqrt(x))/(1-9x^(3))`D. `(3x)/(1-9x^(3))` |
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Answer» We have `y=tan^(-1)((6xsqrt(x))/(1-9x^(3)))` `=tan^(-1)((2cdot(3x^(3//2)))/(1-(3x^(3//2))^(2)))=2 tan ^(-1)(3x^(3//2))` `therefore" "(dy)/(dx)=2xx(1)/(1+9x^(3))xx3xx(3)/(2)xxx^(1//2)=(9sqrt(x))/(1+9x^(3))` `therefore" "g(x)=(9)/(1+9x^(3))` |
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| 9. |
If y=c`e^(x//(x-a))`, then `(dy)/(dx)` equalsA. `a(x-a)^(2)`B. `-(ay)/((x-a)^(2))`C. `a^(2)(x-a)^(2)`D. `a(x-a)` |
| Answer» Correct Answer - B | |
| 10. |
If `f(theta) = sin(tan^(-1)(sintheta/sqrt(cos2theta)))`, where `-pi/4 lt theta lt pi/4,` then the value of `d/(d(tantheta)) f(theta)` is |
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Answer» `f(theta)=sin (tan^(-1)((sin theta)/(sqrt(cos 2theta))))," where "theta in (-(pi)/(4),(pi)/(4))` `=sin (tan^(-1)((sin theta)/(sqrt(2 cos^(2) theta-1))))=sin (sin^(-1)(tan theta))=tan theta` `therefore" "(d(tan theta))/(d(tan theta))=1` |
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| 11. |
`sqrt(x+y)+sqrt(y-x)=c , then (d^2y)/(dx^2)` equalsA. `2//c`B. `-2//c^(2)`C. `2//c^(2)`D. `-2//c` |
| Answer» Correct Answer - C | |
| 12. |
If `sqrt(1-x^6)+sqrt(1-x^6)=a(x^3-y^3),`then prove that `(dy)/(dx)=(x^2)/(y^2)sqrt((1-y^6)/(1-x^6))`A. `(y)/(x)`B. `(x^(2))/(y^(2))`C. `(2y^(2))/(x^(2))`D. `(y^(2))/(x^(2))` |
| Answer» Correct Answer - B | |
| 13. |
If `sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y)`, then `(dy)/(dx)` equalsA. `sqrt((1-x^(2))(1-y^(2)))`B. `sqrt((1-y^(2))/(1-x^(2)))`C. `sqrt((1-x^(2))/(1-y^(2)))`D. none of these |
| Answer» Correct Answer - B | |
| 14. |
If `x^2+y^2=(t+1/t)` and `x^4+y^4=t^2+1/t^2`, then `x^3y(dy)/(dx)=` |
| Answer» Correct Answer - B | |
| 15. |
If `y=int_(0)^(x) f(t)sin{k(x-t)}` dt, then `(d^(2)y)/(dx^(2))+k^(2)` y equals |
| Answer» Correct Answer - C | |
| 16. |
`y= tan^(-1)(sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))` then `dy/dx`A. `(1)/(sqrt(1-x^(4)))`B. `-(1)/(sqrt(1-x^(4)))`C. `(x)/(sqrt(1-x^(4)))`D. `-(x)/(sqrt(1-x^(4)))` |
| Answer» Correct Answer - D | |
| 17. |
`tan^(-1)((x)/(sqrt(1-x)))` |
| Answer» Correct Answer - `(1)/(sqrt(1-x^(2)))` | |
| 18. |
If `y=sqrt(x+sqrt(y+sqrt(x+sqrt(y+...oo))))`, then `(dy)/(dx)` is equal toA. `(y+x)/(y^(2)-2)`B. `(y^(3)-x)/(2y^(2)-2xy-1)`C. `(y^(3)+x)/(2y^(2)-x)`D. none of these |
| Answer» Correct Answer - D | |
| 19. |
If `y=x^(x^(x^(x...^(oo))))` , then `x(1-ylogx)(dy)/(dx)`A. `x^(2)`B. `y^(2)`C. `xy^(2)`D. `xy` |
| Answer» Correct Answer - B | |
| 20. |
`tan^(-1)((1)/(1+sqrt(1-x^(2))))` |
| Answer» Correct Answer - `(1)/(2sqrt(1-x^(2)))` | |
| 21. |
If variables x and y are related by the equation `x=int_(0)^(y)(1)/(sqrt(1+9u^(2)))` du, then `(dy)/(dx)` is equal toA. `(1)/(sqrt(1+9y^(2)))`B. `sqrt(1+9y^(2))`C. `1+9y^(2)`D. `(1)/(1+9y^(2))` |
| Answer» Correct Answer - B | |
| 22. |
Find `(dy)/(dx)`, when: `e^(x)logy=sin^(-1)x+sin^(-1)y` |
| Answer» Correct Answer - `y.sqrt((1-y^(2))/(1-x^(2))).{(1-e^(x)logy.sqrt(1-x^(2)))/((e^(x)sqrt(1-y^(2)))-y)}` | |
| 23. |
If `y=x+(1)/(x+(1)/(x+(1)/(x+...oo)))," prove that "(dy)/(dx)=(y)/((2y-x)).` |
| Answer» Correct Answer - `y=x+(1)/(y).` | |
| 24. |
`d/(dx)[tan^(-1)((sqrt(x)(3-x))/(1-3x))]=``1/(2(1+x)sqrt(x))`(b) `3/((1+x)sqrt(x))``2/((1+x)sqrt(x))`(d) `3/(2(1+x)sqrt(x))`A. `(1)/(2(1+x)sqrt(x))`B. `(3)/((1+x)sqrt(x))`C. `(2)/((1+x)sqrt(x))`D. `(3)/(2(1+x)sqrt(x))` |
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Answer» `(d)/(dx)(tan^(-1)""(sqrt(x)(3-x))/(1-3x))` `(d)/(dx)(tan^(-1)""((tan theta(3-tan^(2)theta)))/(1-3tan^(2)theta))` `("putting "sqrt(x)=tan theta or theta=tan^(-1)sqrt(x))` `=(d)/(dx)(tan^(-1)""((3tantheta-tan^(3)theta))/(1-3tan^(2)theta))` `=(d)/(dx)[tan^(-1)(tan 3theta)]=(d)/(dx)(3theta)` `=(d)/(dx)[3tan^(-1)sqrt(x)]=(3)/(2sqrt(x)(1+x))` |
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| 25. |
If `y=(sin^(-1)x)/(sqrt(1-x^2)),t h e n((1-x^2)dy)/(dx)`is equal to`x+y`(b) `1+x y``1-x y`(d) `x y-2`A. x+yB. 1+xyC. 1-xyD. xy-2 |
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Answer» `y=(sin^(-1))/(sqrt(1-x^(2)))` `therefore" "(dy)/(dx)=(sqrt(1-x^(2))(1)/(sqrt(1-x^(2)))-(sin^(-1)x)(1)/(2)((-2x))/(sqrt(1-x^(2))))/(1-x^(2))` `"or "(1-x^(2))(dy)/(dx)=1+x((sin^(-1)x)/(sqrt(1-x^(2))))=1+xy` |
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| 26. |
`"If "y=a^(x^(a^(x...oo)))", prove that "(dy)/(dx)=(y^(2)(logy))/(x[1-y(logx)(logy)]).` |
| Answer» `y=a^((x^(y)))rArr log y=x^(y)(loga) rArr log(logy)=(ylog x)+log(loga).` | |
| 27. |
If `y=e^(sin^(-1)x)" and "u=logx," then"(dy)/(du),` isA. `(e^(sin^(-1)x))/(sqrt(1-x^(2)))`B. `xe^(sin^(-1)x)`C. `(xe^(sin^(-1)x))/(sqrt(1-x^(2)))`D. `(e^(sin^(-1)x))/(x)` |
| Answer» Correct Answer - C | |
| 28. |
`" If "y=sqrt(logx+sqrt(logx+sqrt(logx+...oo)))," then "(dy)/(dx)` isA. `(x)/(2y-1)`B. `(x)/(2y+1)`C. `(1)/(x(2y-1)`D. `(1)/(x(1-2y)` |
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Answer» `y=sqrt(log x+y)` `"or "y^(2)=log x+y` `"or "2y(dy)/(dx)=(1)/(x)+(dy)/(dx)or (dy)/(dx)=(1)/(x(2y-1))` |
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| 29. |
Find `(dy)/(dx)`, when: `y=(sinx)^(x)+sin^(-1)sqrtx` |
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Answer» Correct Answer - `(sinx)^(x).{x cot x+log sin x}+(1)/(2sqrt(x-x^(2)))` Let `y=u+v,` where `u=(sinx)^(x) and v=sin^(-1)sqrtx` Now, `u=(sinx)^(x) rArr logu=x log sin x` `rArr (1)/(u).(du)/(dx)=x.(cosx)/(sinx)+log sin x.1` `rArr (du)/(dx)=(sinx)^(x).{x cot +log sin x}.` And, `v=sin^(-1)sqrtxrArr(dv)/(dx)=(1)/(sqrt(1-x)).(1)/(2)x^(-1//2)=(1)/(2sqrt(x-x^(2)))` `therefore(dy)/(dx)=(du)/(dx)+(dv)/(dx).` |
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| 30. |
`"If "y=(sinx)^(x)+sin^(-1)sqrtx", find "(dy)/(dx).` |
| Answer» Correct Answer - `(sinx)^(x)[x cot x+log (sinx)]+(1)/(2sqrt(x-x^(2)))` | |
| 31. |
If `y=(sinx)^(x)+sin^(-1)sqrtx,` find `(dy)/(dx).` |
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Answer» Let `(sinx)^(x)=u and sin^(-1)sqrtx=v`. Then, `y=u+v rArr (dy)/(dx)=(du)/(dx)+(dv)/(dx)." …(i)"` Now, `u=(sinx)^(x) rArr logu = xlog (sinx)` `rArr (1)/(u).(du)/(dx)=x.(1)/(sinx).cosx+log(sinx).1` `rArr (du)/(dx)=u{x cot x+log(sinx)}` `rArr (du)/(dx)=(sinx)^(x).{x cot x+log(sinx)}." ....(ii)"` `v=sin^(-1)sqrtxrArr(dv)/(dx)=(1)/(sqrt((1-x))).(1)/(2)x^(-1//2)=(1)/(2sqrtxsqrt((1-x)))=(1)/(2sqrt(x-x^(2)))." ...(iii)"` Using (ii) and (iii) in (i), we get `(dy)/(dx)=(sinx)^(x){x cot x+log(sinx)}+(1)/(2sqrt(x-x^(2))).` |
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| 32. |
If `y = x^(x ^(x^(x ^(x.... oo)` then prove that ` x dy/dx = y^2/(1- y logx )` |
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Answer» We know that an infinite series is not affected by the exclusion of a single term. So, we may write the given function as `y=x^(y).` Now, `y=x^(y)rArr logy=y log x`. On differentiating both sides of (i) w.r.t. x, we get `(1)/(y).(dy)/(dx)=y.(1)/(x)+logx.(dy)/(dx)` `rArr((1)/(y)-logx)(dy)/(dx)=(y)/(x)` `rArr((1-ylogx))/(y).(dy)/(dx)=(y)/(x)` `(dy)/(dx)={(y)/(x)xx(y)/((1-y logx))}rArr (dy)/(dx)=(y^(2))/(x(1-ylogx)).` |
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| 33. |
If `(x^2+y^2)^2=x y`, find `(dy)/(dx)` |
| Answer» Correct Answer - `(y(3x^(2)-y^(2)))/(x(x^(2)-3y^(2)))` | |
| 34. |
If `(tan^(-1)x)^(y)+y^(cotx)=1,` then find `(dy)/(dx).` |
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Answer» Let `(tan^(-1)x)^(y)=u and y^(cotx)=v`. Then, `u+v=1 rArr (du)/(dx)+(dv)/(dx)=0." …(i)"` Now, `u=(tan^(-1)x)^(y)` `rArr logu=ylog (tan^(-1)x)` `rArr(1)/(u).(du)/(dx)=y.(1)/(tan^(-1)x).(1)/((1+x^(2)))+log(tan^(-1)x)(dy)/(dx)` `rArr (du)/(dx)=u.{(y)/((1+x^(2)tan^(-1)x))+log(tan^(-1)x).(dy)/(dx)}` `rArr(du)/(dx)=(tan^(-1)x)^(y).{(y)/((1+x^(2))tan^(-1)x)+log(tan^(-1)x).(dy)/(dx)}` `rArr (du)/(dx)=(y(tan^(-1)x)^(y-1))/((1+x^(2)))+(tan^(-1)x)^(y)log(tan^(-1)x).(dy)/(dx)." ...(ii)"` Again, `v=y^(cotx)` `rArr logv=(cotx)logy` `rArr (1)/(v).(dv)/(dx)=(cotx).(1)/(y).(dy)/(dx)+(logy)(-"cosec"^(2)x)` `rArr(dv)/(dx)=v{(cotx)/(y).(dy)/(dx)-("cosec"^(2)x)(logy)}` `rArr (dv)/(dx)=y^(cotx){(cotx)/(y).(dy)/(dx)-("cosec"^(2)x)(logy)}` `rArr (dv)/(dx)=(cotx)y^((cotx)-1).(dy)/(dx)-y^(cotx)("cosec"^(2)x)(logy)." ...(iii)"` Using (ii) and (iii) in (i), we get `(y(tan^(-1)x)^(y-1))/((1+x^(2)))+(tan^(-1)x)^(y)log(tan^(-1)x)(dy)/(dx)+(cotx)y^((cotx)-1).(dy)/(dx)-y^(cotx)("cosec"^(2)x)(logy)=0` `rArr{(tan^(-1)x)^(y)log(tan^(-1)x)+(cotx)y^((cotx)-1)}(dy)/(dx)={y^(cotx)("cosec"^(2)x)(logy)-(y(tan^(-1)x)^(y-1))/((1+x^(2)))}` `rArr(dy)/(dx)=(y^(cotx)("cosec"^(2)x)(logy)-(y(tan^(-1)x)^(y-1))/((1+x^(2))))/((tan^(-1)x)^(y)log(tan^(-1)x)+(cotx)y^((cotx)-1)).` |
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| 35. |
If `y=x^(cotx)+(2x^(2)-3)/(x^(2)+x+2)`, find `(dy)/(dx).` |
| Answer» Correct Answer - `x^(cotx)((cotx)/(x)-"cosec"^(2)x.logx)+(2x^(2)+14x+3)/((x^(2)+x+2)^(2)).` | |
| 36. |
Find `(dy)/(dx)`, when: `y=(tanx)^(cotx)` |
| Answer» Correct Answer - `(tanx)^(cotx).{"cosec"^(2)x(1-log tanx)}` | |
| 37. |
Differentiate each of the following w.r.t. x : (i) `sin (logx), x gt0` (ii) `log(logx), xgt1` |
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Answer» (i) Let `y=sin(logx)`. Putting `logx=t,` we get `y=sin t and t=logx` `rArr(dy)/(dx)=cost and (dt)/(dx)=(1)/(x)` `rArr(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(costxx(1)/(x))=cos(logx)xx(1)/(x)=(cos(logx))/(x)`. Hence, `(d)/(dx){sin(logx)}=(cos(logx))/(x).` (ii) Let `y=log(logx)`. Putting `logx=t,` we get `y=logt and t=log x` `rArr(dy)/(dx)=(1)/(t)and (dt)/(dx)=(1)/(x)` `rArr(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=((1)/(t)xx(1)/(x))=((1)/(logx)xx(1)/(x))=(1)/((xlogx))`. `therefore(d)/(dx){log(logx)}=(1)/((xlogx)).` |
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| 38. |
`sqrt(1+cotx)` |
| Answer» Correct Answer - `-(1)/(2).("cosec"^(2)x)/(sqrt(1+cotx))` | |
| 39. |
If `y=e^(sqrt(cotx))`, find `(dy)/(dx)`. |
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Answer» Given: `y=e^(sqrt(cotx))`. Putting `cotx=t and sqrt(cotx)=sqrt(t)=u`, we get `y=e^(u),u=sqrt(t)and t=cotx` `rArr(dy)/(dx)=e^(u),(du)/(dt)=(1)/(2)t^(-1//2)=(1)/(2sqrtt)and (dt)/(dx)=-"cosec"^(2)x` `rArr(dy)/(dx)=((dy)/(du)xx(du)/(dt)xx(dt)/(dx))` `={e^(u).(1)/(2sqrtt).(-"cosec"^(2)x)}=e^(sqrt(cotx)).(1)/(2sqrt(cotx)).(-"cosec"^(2)x)` `=((-"cosec"^(2)x)e^(sqrt(cotx)))/(2sqrt(cotx)).` |
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| 40. |
If `y=log tan.(x)/(2)`, find `(dy)/(dx)`. |
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Answer» Given: `y=log tan.(x)/(2)`. Putting `(x)/(2)=t and tan.(x)/(2)=tant=u`, we get `y=logu, u = tan t and t=(x)/(2)` `rArr(dy)/(dx)=(1)/(u),(du)/(dt)=sec^(2)t and (dt)/(dx)=(1)/(2)` `rArr(dy)/(dx)=((dy)/(du)xx(du)/(dt)xx(dt)/(dx))` `=((1)/(u)xxsec^(2)txx(1)/(2))=((1)/(2tant)xxsec^(2)t)=(1)/(sin2t)=(1)/(sinx)="cosec x".` Hence, `(dy)/(dx)="cosec x"`. |
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| 41. |
If `y=sqrt(e^(sqrtx))`, find `(dy)/(dx)`. |
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Answer» Putting `sqrtx=t, e^(sqrtx)=e^(t)=u`, we get `y=sqrtu, u=e^(t) and t= sqrtx` `rArr(dy)/(dx)=(1)/(2)u^(-1//2)=(1)/(2sqrtu),(du)/(dt)=e^(t)and (dt)/(dx)=(1)/(2)x^(-1//2)=(1)/(2sqrtx)` `rArr(dy)/(dx)=((dy)/(du)xx(du)/(dt)xx(dt)/(dx))` `=((1)/(2sqrtu)xxe^(t)xx(1)/(2sqrtx))={(1)/(2sqrtu)xxuxx(1)/(2sqrtx)}=(sqrtu)/(4sqrtx)=(e^((1)/(2)t))/(4sqrtx)` `=(e^((1)/(2)sqrtx))/(4sqrtx)`. |
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| 42. |
If `y=(1)/(logcosx)`, find `(dy)/(dx).` |
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Answer» Given : `y=(log cos x)^(-1).` Putting `cosx=t` and `log cos x=logt=u`, we get `y=u^(-1)=(1)/(u),u=log t and t = cos x` `rArr(dy)/(dx)=(-1)/(u^(2)),(du)/(dt)=(1)/(t) and (dt)/(dx)=-sinx` `rArr(dy)/(dx)=((dy)/(dx)xx(du)/(dt)xx(dt)/(dx))` `={(-1)/(u^(2))xx(1)/(t)xx(-sinx)}={(1)/((log cos x)^(2)).(1)/(cos x).sinx}` `=(tanx)/((log cos x)^(2))` |
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| 43. |
If `x=e^(cos2t)`and `y=e^(sin2t)`, prove that `(dy)/(dx)=-(ylogx)/(xlogy)` |
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Answer» `x=e^(cos2t)and y=e^(sin 2t)` ` cos 2t= log x and sin 2t = log y` `therefore" "cos^(2) 2t +sin^(2) 2t = (log x)^(2) + (log y)^(2)` `rArr" "(log x)^(2)+(log y)^(2)=1` Differentiating both sides w.r.t. x, we get `2log x(1)/(x)+2 log y (1)/(y)(dy)/(dx)=0` `rArr" "(dy)/(dx)=(-y log x)/(x log y)` |
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| 44. |
Find `(dy)/(dx)`, when: `sin^(2)x+2cosy+xy=0` |
| Answer» Correct Answer - `((y+sin2x))/((2siny-x))` | |
| 45. |
Find the derivative of following functions w.r.t. x: sin {cos (x2)} |
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Answer» Let y = sin {cos (x2)} Diff. w.r.t. x, dy/dx= d/dx sin (cos(x2)) cos (cos x2). (- sin x2) d/dxx2 = -cos (cos x2) sin (x2). 2x = -2x sin (x2) cos (cos x2) |
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| 46. |
Differentiatethe following functions with respect to`tan(x^0+45^0)` |
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Answer» 3)`y=tan(x^0+45^0)` `=tan(pi/180(x^0+45^0))` diff with respect to x `dy/dx=sec^2[pi/180(x^0+45^0]]*d/dx(pi/180(x^0+45^0))` `pi/180*sec^2[pi/180(x^0+45^0)]`. |
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| 47. |
Differentiate with respect to x:(3 – 4x)5 |
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Answer» Chain rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx(3 – 4x)5 = 5(3 – 4x)4 * d/dx(3 – 4x) = 5(3 – 4x)4 * ( – 4) = – 20(3 – 4x)4 |
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| 48. |
Differentiate the following functions with respect to `x`:`tan^(-1){sqrt((1+cosx)/(1-cosx))}` |
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Answer» `sqrt((1+cosx)/(1-cosx)) = sqrt((2cos^2(x/2))/(2sin^2(x/2)))` `=cot(x/2)` Let `y = tan^-1(sqrt((1+cosx)/(1-cosx)) )` `=>y = tan^-1(cot(x/2))` `=>y = pi/2 - cot^-1(cot(x/2))` `=>y = pi/2 - x/2` Differentiating both sides w.r.t. `x`, `=>dy/dx = 0-1/2` `:. dy/dx =-1/2.` |
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| 49. |
Differentiate with respect to x:(5 + 7x)6 |
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Answer» Chain rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx(5 + 7x)6 = 6(5 + 7x)5 * d/dx(5 + 7x) = 6(5 + 7x)5 * 7 = 42(5 + 7x)5 |
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| 50. |
If `y=cos^(-1)(2x)+2cos^(-1)sqrt(1-4x^2), -1/2 |
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Answer» Correct Answer - `(2)/(sqrt(1-4x^(2)))` Put `2x=cos theta.` |
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