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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.
| 301. |
Find `(dy)/(dx)`for `y=sin(x^2+1)dot` |
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Answer» `"Let "y= sin(x^(2)+1).` Putting `u=x^(2)+1," we get y sin u"` `therefore" "(dy)/(dx)=cos u and (du)/(dx) = 2x` `"Now, "(du)/(dx)=(dy)/(du).(du)/(dx)` `= (cos u) (2x)=2x cos(x^(2)+1)` |
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| 302. |
If `y=sin^(-1)[xsqrt(1-x)-sqrt(x)sqrt(1-x^2])`and `0 |
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Answer» `y=sin^(-1)[x sqrt(1-x)-sqrt(x)sqrt(1-x^(2))]," where "0lt x lt 1` `=sin^(-1)[xsqrt(1-(sqrt(x))^(2))-sqrt(x)sqrt(1-x^(2))]` `=sin^(-1)x-sin^(-1)sqrt(x)` `[because sin^(-1) x-sin^(-1)y=sin^(-1)(xsqrt(1-y^(2))-ysqrt(1-x^(2)))]` Differentiating w.r.t.x, we get `(dy)/(dx)=(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-(sqrt(x))^(2)))(d)/(dx)(sqrt(x))` `=(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-x))xx(1)/(2sqrt(x))` |
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| 303. |
`y=tan^(-1)""(3x-x^(3))/(2x^(2)-1),-(1)/(sqrt(3))ltxlt(1)/(sqrt(3))` |
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Answer» Correct Answer - `(3)/(1+x^(2))` `y=tan^(-1)((3x-x^(3))/(1-3x^(2)))` `Put x tan theta.` Then, `y=tan^(-1)((3 tan theta - tan^(3)theta)/(1-3 tan^(2)theta))` `=tan^(-1)(tan 3theta)` `=3theta` `=3tan^(-1)x` `therefore" "(dy)/(dx)=(3)/(1+x^(2))` |
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| 304. |
If `y=(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))` then `(dy)/(dx)` at `x=0` is |
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Answer» Correct Answer - 1 `(dy)/(dx)=1(1+x^(2))(1+x^(4))...(1+x^(2^(n)))+2x(1+x)(1+x^(4))...(1+x^(2^(n)))` `+4x^(3)(1+x)(1+x^(2))(1+x^(8))...(1+x^(2^(n)))` : `+2^(n)x^(2^(n-1))(1+x)(1+x^(2))...(1+x^(2^(n-1)))` `"or "(dy)/(dx):|_(x=0)=1` |
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| 305. |
If `y=e^(x) log (sin 2x),` find `(dy)/(dx)`. |
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Answer» We have `(dy)/(dx)=(d)/(dx)[e^(x)log(sin 2x)]` `e^(x).(d)/(dx){log(sin 2x)}+log(sin 2x).(d)/(dx)(e^(x))` `=e^(x).{(1)/(sin 2x).cos2x.2}+log(sin 2x).e^(x)` `=2e^(x) cot 2x+e^(x) log (sin 2x)` `=e^(x){2 cot 2x+log (sin 2x)}`. |
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| 306. |
Find `(dy)/(dx)`for `y=tan^(-1)sqrt((a-x)/(a+x),)-a |
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Answer» `y=tan^(-1){sqrt((a-x)/(a+x))}," where " -a lt x lt a` Substituting `x= a cos theta,` we get `y=tan^(-1){sqrt((a-a cos theta)/(a+a cos theta))},` `=tan^(-1){sqrt((1- cos theta)/(1+ cos theta))},` `=tan^(-1){sqrt(tan^(2)""(theta)/(2))}` `=tan^(-1)|tan""(theta)/(2)|` Also, for `-a lt x lt a, -1 lt cos theta lt 1` `"or "theta in (0,pi) or (theta)/(2) in (0, (pi)/(2))` `therefore" "y=tan^(-1)|tan""(theta)/(2)|=tan^(-1)(tan""(theta)/(2))` `=(theta)/(2)=(1)/(2) cos^(-1) ((x)/(a))` `"or "(dy)/(dx)=-(1)/(2)xx(1)/(sqrt(1-(x^(2))/(a^(2))))(d)/(dx)((x)/(a))=-(1)/(2sqrt(a^(2)-x^(2)))` |
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| 307. |
Using the definition of derivative find the derivative of `sqrt(sin x)` |
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Answer» Correct Answer - `(cos x)/(2sqrt(sin x))` (b) `-sin 2x` (c) `(1)/(sqrt(1-x^(2)))` (a)`"Let "f(x)=sqrt(sin x)` `therefore" "f(x+h)=sqrt(sin(x+h))` `therefore" "(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)` `=underset(hrarr0)lim(sqrt(sin(x+h))-sqrt(sin x))/(h)` `=underset(hrarr0)lim(sin (x+h)-sin x)/(h(sqrt(sin(x+h)+sqrt(sinx))))` `=underset(hrarr0)lim(2sin""((h)/(2))cos""((2x+h)/(2)))/(h(sqrt(sin(x+h)+sqrt(sin x))))` `=underset(hrarr0)lim((sin""(h)/(2)))/(((h)/(2)))underset(hrarr0)lim(cos(x+(h)/(2)))/((sqrt(sin (x_+h)+sqrt(sin x))))` `=(cos x)/(sqrt(sin x )+sqrt(sin x))=(cos x)/(2sqrt(sin x))` (b) Let `f(x) =cos^(2)x.` `therefore" "f(x+h)=cos^(2)(x+h)` `therefore" "(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)` `=underset(hrarr0)lim(cos^(2)(x+h)-cos^(2)x)/(h)` `=underset(hrarr0)lim(sin^(2)x-sin^(2)(x+h))/(h)` `=underset(hrarr0)lim(sin (x+x+h)sin (x-(x+h)))/(h)` `=underset(hrarr0)lim(sin(2x+h)sin(-h))/(h)` `=-underset(hrarr0)lim(sin h)/(h).underset(hrarr0)limsin(2x+h)` `=-sin 2x` (c) `(d(sin^(-1)x))/(dx)=underset(hrarr0)lim(sin^(-1)(x+h)-sin^(-1)x)/(h)` `=underset(hrarr0)lim(sin^(-1)[(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])/(h)` `=underset(hrarr0)lim(sin^(-1)[(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])/([(x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^(2))])` `xx((x+h)sqrt(1-x^(2))-xsqrt(1-(x+h)^2))/(h)` `=1xxunderset(hrarr0)lim((x+h)^(2)(1-x^(2))-x^(2)(1-(x+h)^(2)))/(h[(x+h)sqrt(1-x^(2))+xsqrt(1-(x+h)^(2))])` `=underset(hrarr0)lim((x+h)^(2)-x)/(h[(x+0)sqrt(1-x^(2))+xsqrt(1-(x+0)^(2))])` `=underset(hrarr0)lim(h(2x+h))/(h[2xsqrt(1-x^(2))])` `=underset(hrarr0)lim(2x+h)/(2xsqrt(1-x^(2)))` `=(1)/(sqrt(1-x^(2)))` |
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| 308. |
If `y=sin^(-1)((2x)/(1+x^2)),`then find `(dy)/(dx)` |
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Answer» Correct Answer - `(2)/(1+x^(2))` `y=sin^(-1)((2x)/(1+x^(2)))` Let `x = tan theta.` Therefore, `y=sin^(-1)((2tan theta)/(1+tan^(2)theta))` `=sin^(-1)(sin 2 theta)` `=2theta` `=2tan^(-1)x` `therefore" "(dy)/(dx=(d)/(dx)(2 tan^(-1)x)` `=2(d)/(dx)(tan^(-1)x)` `(2)/(1+x^(2))` |
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| 309. |
`"If "x^(3)+y^(3)=3 axy," find "(dy)/(dx).` |
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Answer» Correct Answer - `(ay-x^(2))/(y^(2)-ax)` Given `x^(3)+y^(3)-3axy = 0`. From `(dy)/(dx)=-("differentiating f w.r.t. x keeping y as consant")/("differentiating f w.r.t. y keeping x as constant")` `=-(3x^(2)-3ay)/(3y^(2)-3ax)=(ay-x^(2))/(y^(2)-ax)` |
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| 310. |
Find `(dy)/(dx),`when`x=(3a t)/(a+t^2)"and"y=(3a t^2)/(1+t^2)` |
| Answer» Correct Answer - `(2t)/((1-t^(2)))` | |
| 311. |
If`y=sqrt(log{sin((x^2)/3-1)}),t h e nfin d(dy)/(dx)dot` |
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Answer» `y=sqrt(log{sin((x^(3))/(3)-1)}}` Putting `(x^(3))/(3)-1=v,` we get sin `((x^(2))/(3)-1)` = sin v = u. Putting log `{sin((x^(2))/(3)-1)} = log y = z, we get y =sqrt(z),z = log u.` `u=sin v, and v=(x^(2))/(3)-1.` Therefore, `(dy)/(dx)(1)(2sqrt(z)), (dz)/(du) =(1)/(u), (du)/(dv) = cos v, and (dv)/(dx)=(2x)/(3)` `Now,(dy)/(dx)=(dy)/(dz)xx(dz)/(du)xx(du)/(dv)xx(dv)/(dx)` `=((1)/(2sqrt(z))((1)/(u))(cos v) ((2x)/(3))` `=(x)/(3).(cos v)/(usqrt(log u))` `=(x cot((x^(2))/(3)1))/(3sqrt(log{sin((x^(2))/(3)-1)}})` |
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| 312. |
If `x y=e^((x-y)),`then find `(dy)/(dx)` |
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Answer» Correct Answer - `(y(x-1))/(x(y+1))` The given function is `xy=e^((x-y)).` Taking logarithm on both the sides, we obtain `log(xy)=log (e^(x-y))` `log x + log y = (x-y)` Differentiating both sides with respect to x, we get `(1)/(x)+(1)/(y)(dy)/(dx)=1-(dy)/(dx)` `"or "(1+(1)/(y))(dy)/(dx)=1-(1)/(x)` `therefore" "(dy)/(dx)=(y(x-1))/(x(y+1))` |
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| 313. |
`y=tan^(-1)((sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))),w h e r e``-1 |
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Answer» Putting `x^(2)= cos 2theta,` we get `y=tan^(-1)((sqrt(1+cos theta)+sqrt(1-cos theta))/(sqrt(1+cost 2theta)-sqrt(1-cost 2theta)))` `=tan^(-1)((sqrt(2cos^(2) theta)+sqrt(2sin^(2) theta))/(sqrt(2cos^(2)theta)-sqrt(2 sin^(2)theta)))` `=tan^(-1)((cos theta+sintheta)/(cos theta-sin theta))` `=tan^(-1)((1+tan theta)/(1- tan theta))` `=tan^(-1)(tan (pi//4 +theta))` `[{:(because, 0lt x^(2) lt1 rArr0 lt cos 2 theta lt 1),(or ,0 lt 2theta lt pi//2), (or , 0 lt theta lt pi//4) , (or, pi//4 lt pi//4 + theta lt pi//2):}]` `=(pi)/(4)+theta` `(pi)/(4)+(1)/(2)cos^(-1) x^(2)` `therefore" "(dy)/(dx)=0-(1)/(2)xx(1)/(sqrt(1-(x^(2))^(2))).(d)/(dx)(x^(2))` `-(1)/(2)xx(2x)/(sqrt(1-x^(4)))=(-x)/(sqrt(1-x^(4)))` |
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| 314. |
If `x^(y)=y^(x)`, fin `(dy)/(dx).` |
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Answer» Given : `x^(y)=y^(x)` `rArr ylog x=x logy." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `y.(d)/(dx)(logx)+(logx).(d)/(dx)(y)=x.(d)/(dx)(logy)+(logy).(d)/(dx)(x)` `rArry.(1)/(x)+(logx).(dy)/(dx)=x.(1)/(y).(dy)/(dx)=x.(1)/(y).(dy)/(dx)+(logy).1` `rArr(logx-(x)/(y))(dy)/(dx)=(logy-(y)/(x))` `rArr((ylogx-x))/(y).(dy)/(dx)=((xlogy-y))/(x)` `rArr(dy)/(dx)=(y(xlogy-y))/(x(logx-x)).` |
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| 315. |
Find the derivative of `sqrt(4-x)`w.r.t. `x`using the first principle. |
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Answer» Correct Answer - `(-1)/(2sqrt(4-x))` `"Let "f(x)=sqrt(4-x)." Then "f(x+h)=sqrt(4-(x+h))." Therefore,"` `(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)` `=underset(hrarr0)lim(sqrt(4-(x+h))-sqrt(4-x))/(h)` `=underset(hrarr0)lim({sqrt(4-(x+h))-sqrt(4-x)}{sqrt(4-(x+h))+sqrt(4-x)})/(h{sqrt(4-(x+h))+sqrt(4-x)})` `=underset(hrarr0)lim(4-(x+h)-(4-x))/(h{sqrt(4-(x+h))+sqrt(4-x)})` `=underset(hrarr0)lim(-h)/(h{sqrt(4-x-h)+sqrt(4-x)})=(-1)/(2sqrt(4-x))` |
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| 316. |
If `y=(1+x^(1/4))(1+x^(1/2))(1-x^(1/4))`, then find `(dy)/(dx)dot` |
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Answer» `y=(1+x^(1//4))(1+x^(1//2))(1-x^(1//4))` `=(1+x^(1//4))(1-x^(1//4))(1+x^(1//2))` `=(1-x^(1//2))(1+x^(1//2))` `=1-x` `therefore" "(dy)/(dx)=-1` |
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| 317. |
If `x^(2)+y^(2)=a^(2)`, then the value of `(dy)/(dx)` is -A. `(x)/(y)`B. `-(x)/(y)`C. `(y)/(x)`D. `-(y)/(x)` |
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Answer» Correct Answer - B |
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| 318. |
If `y^x=x^y ,t h e nfin d(dy)/(dx)dot` |
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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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| 319. |
`"Find "(dy)/(dx)" for "y=log(x+sqrt(a^(2)+x^(2))).` |
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Answer» `y=log (x+sqrt(a^(2)+x^(2)))` `"Then "(dy)/(dx)=(d)/(x){log (x+sqrt(a^(2)+x^(2)))}` `=(1)/(x+sqrt(a^(2)+x^(2)))(d)/(dx)(x+sqrt(a^(2)+x^(2)))` `=(1)/(x+sqrt(a^(2)+x^(2)))xx{1+(1)/(2)(a^(2)+x^(2))^(-1//2)(d)/(dx)(a^(2)+x^(2))}` `=(1)/(x+sqrt(a^(2)+x^(2))){1+(1)/(2sqrt(a^(2)+x^(2)))xx2x}` `(1)/(x+sqrt(a^(2)+x^(2)))xx(sqrt(a^(2)+x^(2))+x)/(sqrt(a^(2)+x^(2)))` `=(1)/(sqrt(a^(2)+x^(2)))` |
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| 320. |
If `y=log sin (e^(x)+5x+8)`, find `(dy)/(dx).` |
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Answer» Given `y=log sin (e^(x)+5x+8)`. On differentiating both sides of (i) w.r.t. x, we get `(dy)/(dx)=(1)/((e^(x)+5x+8)).cos(e^(x)+5x+8).(d)/(dx)(e^(x)+5x+8)` `={cot(e^(x)+5x+8)}(e^(x)+5)=(e^(x)+5).cot(e^(x)+5x+8).` |
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| 321. |
If `y=btan^(-1)(x/a+tan^(-1)y/x),fin d(dy)/(dx)dot` |
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Answer» Correct Answer - `((1)/(a)-(y)/(x^(2)+y^(2)))/((1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2)))` We have `y=b tan^(-1)((x)/(a)+tan^(-1)""(y)/(x))` `"or "tan""(y)/(b)=(x)/(a)+tan^(-1)""(y)/(x)` Differentiating both sides w.r.t. x, we get `(1)/(b)sec^(2)((y)/(b))(dy)/(dx)=(1)/(a)+(1)/(1+((y)/(x))^(2))xx(x(dy)/(dx)-y)/(x^(2))` `"or "(1)/(b)sec^(2)((y)/(b))(dy)/(dx)=(1)/(a)+(x(dy)/(dx)-y)/(x^(2)+y^(2))` `"or "(dy)/(dx){(1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2))}=(1)/(a)-(y)/(x^(2)+y^(2))` `"or "(dy)/(dx)=((1)/(a)-(y)/(x^(2)+y^(2)))/((1)/(b)sec^(2)((y)/(b))-(x)/(x^(2)+y^(2)))` |
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| 322. |
If `x=a cos theta, y=a sin theta`, then the value of `(dy)/(dx)` is -A. `tan theta`B. `-tan theta`C. `-cot theta`D. `cot theta` |
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Answer» Correct Answer - C |
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| 323. |
Find the derivative of `e^(sqrt(x))`w.r.t. `x`using the first principle. |
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Answer» Let `f(x)=""_(e)sqrt(x)`. Then `f(x+h)=""_(e)sqrt(x+h)` `therefore" "(d)/(dx)(f(x))=underset(hrarr0)lim(f(x+h)-f(x))/(h)` `=underset(hrarr0)lim(""_(e)sqrt(x+h)-""_(e)sqrt(x))/(h)` `=""_(e)sqrt(x)" "underset(hrarr0)lim(""_(e)sqrt(x+h)-sqrt(x)-1)/(h)` `=""_(e)sqrt(x)underset(hrarr0)lim((esqrt(x+h)-sqrt(x)-1)/(sqrt(x+h)-sqrt(x)))((sqrt(x+h)-sqrt(x))/(h))` `=""_(e)sqrt(x)underset(hrarr0)lim((esqrt(x+h)-sqrt(x)-1)/(sqrt(x+h)-sqrt(x)))` `x((sqrt(x+h)-sqrt(x))(sqrt(x+h)+sqrt(x)))/(h(sqrt(x+h)+sqrt(x)))` `=""_(e)sqrt(x)underset(hrarr0)lim((e^(y)-1)/(y))underset(hrarr0)lim((x+h-x)/(h(sqrt(x+h)+sqrt(x))))` `"where "y=sqrt(x+h)-sqrt(x)" "(because" when " hrarr0,yrarr0)` `therefore" "(d)/(dx)(f(x))=""_(e)sqrt(x)xx1xx((1)/(sqrt(x)+sqrt(x)))=(""_(e)sqrt(x))/(2sqrt(x))` |
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| 324. |
`"Find "(dy)/(dx) if x =a(theta- sin theta) and y= a (1- cos theta).` |
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Answer» `"We have "x=a(theta-sin theta)and y=a (1- cos theta).` `therefore" "(dx)/("d"theta)=a(1-cos theta) and (dy)/("d"theta)=a sin theta` `"or "(dy)/(dx)=(dy//"d" theta)/(dx//"d" theta)` `=(a sin theta)/(a(1-cos theta))` `=(2sin (theta//2)cos (theta//2))/(2 sin^(2) (theta//2))` `=cot""(theta)/(2)` |
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| 325. |
If `y=x+e^x ,`then `(d^2x)/(dy^2)`is equal toA. `(1)/((1+e^(x))^(2))`B. `-(e^(x))/((1+e^(x))^(2))`C. `-(e^(x))/((1+e^(x))^(3))`D. `e^(x)` |
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Answer» Correct Answer - C We have, `y=x+e^(x)` `implies" "(dy)/(dx)=1+e^(x)` `implies" "(dx)/(dy)=(1)/(1+e^(2))` `implies" "(d^(2)x)/(dy^(2))=(d)/(dy)((1)/(1+e^(x)))` `implies" "(d^(2)x)/(dy^(2))=-(1)/((1+e^(x))^(2))(d)/(dy)(1+e^(x))` `implies" "(d^(2)x)/(dy^(2))=-(1)/((1+e^(x))^(2))e^(x)(dx)/(dy)=(-e^(x))/((1+e^(x))^(3))` |
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| 326. |
`y=sqrt(x^2+1) - log(1/x+sqrt(1+1/(x^2)))`, find `dy/dx` |
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Answer» We have `y=sqrt(x^(2)+1)-log{(1+sqrt(x^(2)+1))/(x)}` `rArr y=sqrt(x^(2)+1)-log{1+sqrt(x^(2)+1)}+logx` `rArr(dy)/(dx)=(1)/(2)(x^(2)+1)^(-1//2).2x-(1)/({1+sqrt(x^(2)+1)}).{(1)/(2)(x^(2)+1)^(-1//2).2x}+(1)/(x)` `=(x)/(sqrt(x^(2)+1))-(1)/({1+sqrt(x^(2)+1)}).(x)/(sqrt(x^(2)+1))+(1)/(x)` `=(x{1+sqrt(x^(2)+1}}-x)/((sqrt(x^(2)+1)){1+sqrt(x^(2)+1)})+(1)/(x)=(xsqrt(x^(2)+1))/((sqrt(x^(2)+1)){1+sqrt(x^(2)+1)})+(1)/(x)` `=(x)/({1+sqrt(x^(2)+1)})+(1)/(x)=((x^(2)+1)+sqrt(x^(2)+1))/(x{1+sqrt(x^(2)+1)})` `((sqrt(x^(2)+1)){(sqrt(x^(2)+1))+1})/(x{1+sqrt(x^(2)+1)})=(sqrt(x^(2)+1))/(x)`. |
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| 327. |
Find `(dy)/(dx)`, when `x=b s in^2theta`and `y=a cos^2theta` |
| Answer» Correct Answer - `(-b)/(a)` | |
| 328. |
If `x=(2t)/(1+t^2),y=(1-t^2)/(1+t^2),t h e nfin d(dy)/(dx)a tt=2.` |
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Answer» Correct Answer - `(4)/(3)` `(dx)/(dt)=((1+t^(2))2-2txx2t)/((1+t^(2))^(2))=(2-2t^(2))/((1+t^(2))^(2))` `(dy)/(dt)((1+t^(2))(-2t)-(1-t^(2))2t)/((1+t^(2))^(2))=(-4t)/((1+t^(2))^(2))` `therefore" "(dy)/(dx)=(dy//dt)/(dx//dt)=(-4t)/(2-2t^(2))=(2t)/((1+t^(2))^(2))` `"or "(dy)/(dx):|_(t=2)=(4)/(3)` |
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| 329. |
If `x = 3 cos theta - 2 cos^3 theta,y = 3 sin theta - 2 sin^3 theta`, then `dy/dx` is |
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Answer» Correct Answer - `cot theta` `"We have, "x=3 cos theta -2 cos^(3)theta` `"and "y=3 sin theta - 2 sin ^(3)theta` `therefore" "(dx)/(d""theta)=-3 sin theta -2xx3 cos^(2) theta(d)/(d""theta)(cos theta)` `=-3 sin theta + 6 cos^(2) theta sin theta` `"And "(dy)/(d""theta)=3cos theta -2xx3sin^(2)theta(d)/(d""theta)(sin theta)` `=3 cos theta -6 sin^(2) theta. cos theta` `"Now, "(dy)/(dx)=(dy//d""theta)/(dx//d""theta)=(3cos theta-6 sin^(2)theta cos theta)/(-3 sin theta + 6cos^(2) theta sin theta)` `=(3cos theta(1-2 sin^(2)theta))/(3 sin theta(-1+2cos^(2)theta))=cot theta. (cos 2theta)/(cos 2theta)=cot theta` |
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| 330. |
Find the second derivative of sin 3 x cos 5x |
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Answer» y = \(\cfrac12\) [sin(5x + 3x) + sin(5x - 3x)] y = \(\cfrac12\) sin 8x + \(\cfrac12\) sin 2x Differentiating with respect to x \(\cfrac{dy}{d\text x}\) = \(\cfrac82\) cos 8x + \(\cfrac22\) cos 2x ⇒ \(\cfrac{dy}{d\text x}\) = 4 cos 8x + cos 2x Differentiating with respect to x \(\cfrac{d^2y}{d\text x^2}\) = -32 sin 8x - 2 sin 2x Hence Proved |
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| 331. |
The second order derivative of `a sin^3 t ` w.r.t, `a cos^3 t` at `t = pi/4` isA. `(4sqrt(2))/(3a)`B. 2C. `(1)/(12a)`D. 0 |
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Answer» Correct Answer - A Let `y=asin^(3)t" and "x=acos^(3)t`. Then, `(dy)/(dt)=3asin^(2)tcost" and "(dx)/(dt)=-3acos^(2)tsint` `implies" "(dy)/(dx)=-tant` `implies" "(d^(2)y)/(dx^(2))=-sec^(2)t(dt)/(dx)=(sec^(2)t)/(3acos^(2)tsint)` `implies" "((d^(2)y)/(dx^(2)))_(t=pi//4)=(4sqrt(2))/(3a)` |
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| 332. |
If `t=(5^(x))/(x^(5))`, find `(dy)/(dx).` |
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Answer» Given : `y=(5^(x))/(x^(5))." …(i)"` Taking logarithm on both sides of (i), we get `logy=log(5^(x))-log(x^(5))` `rArr log y = x log 5-5log x` `rArr(1)/(y).(dy)/(dx)=(log5).1-(5)/(x)" [differentiating both sides w.r.t.x]"` `rArr(dy)/(dx)=y(log5-(5)/(x))` `rArr(dy)/(dx)=(5^(x))/(x^(5))(log5-(5)/(x)).` |
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| 333. |
Prove that `(d)/(dx)(sin^(-1)x)=(1)/(sqrt(1-x^(2))`, where `x in [-1,1].` |
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Answer» Let `y=sin^(-1)x`, where `x in [-1,1] and y- in [-(pi)/(2),(pi)/(2)].` Then, `y=sin^(-1)x rArr x= sin y` `rArr(dx)/(dy)=cosy ge0" since "yin[-(pi)/(2),(pi)/(2)]` `rArr(dx)/(dy)=sqrt(1-sin^(2)y)=sqrt(1-x^(2))` `rArr(dy)/(dx)=(1)/(sqrt(1-x^(2)))` Hence, `(d)/(dx)(sin^(-1)x)=(1)/(sqrt(1-x^(2))).` |
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| 334. |
If `y=cos^(-1)((2cosx-3sinx)/(sqrt(13)))`, then `(dy)/(dx)`, isA. zeroB. constant=1C. `"constant "ne1`D. none of these |
| Answer» Correct Answer - B | |
| 335. |
Find `(dy)/(dx)` if x= 3 cos theta - cos 2theta and y= sin theta - sin 2theta.` |
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Answer» Correct Answer - `(cos theta -2 cos 2 theta)/(2 sin 2 theta - sin theta)` The given equations are `x=cos theta - cos 2 theta` `and" "y = sin theta - sin 2theta` `"Then, "(dx)/(d""theta)=-sin theta - (-2 sin 2theta)=2 sin 2 theta- sin theta` `"And "(dy)/(d""theta)=cos theta -2 cos 2theta` `therefore" "(dy)/(dx)=(dy//d""theta)/(dx//d""theta)=(cos theta-2 cos 2theta)/(2 sin 2theta- sin theta)` |
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| 336. |
`y=tan^(-1)(x/(1+sqrt(1-x^2)))` |
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Answer» Correct Answer - `(1)/(2sqrt(1-x^(2)))` `y=tan^(-1)""((x)/(1+sqrt(1-x^(2))))` Put `x= sin theta.` Then, `y=tan^(-1)((sintheta)/(1+sqrt(1-sin^(2)theta)))=tan^(-1)((sintheta)/(1+cos theta))` `=tan^(-1)""(2sin""(theta)/(2)cos""(theta)/(2))/(2cos^(2)""(theta)/(2))=tan^(-1)tan""(theta)/(2)=(theta)/(2)` `"So, "y=(sin^(-1)x)/(2)or(dy)/(dx)=(1)/(2sqrt(1-x^(2)))` |
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| 337. |
Find the second derivative of sin `3x cos 5x`. |
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Answer» Correct Answer - `(2sin 2x-32sin 8x)` `2sin A cos B = sin(A+B)+sin (A-B).` |
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| 338. |
Differentiate `x^(x)` w.r.t. x. |
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Answer» Let `y=x^(x)." …(i)"` Taking logarithm on both sides of (i), we get `logy=x log x." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=x.(d)/(dx)(logx)+logx.(d)/(dx)(x)` `=(x.(1)/(x)+logx.1)=(1+logx)` `rArr(dy)/(dx)=y(1+logx)` `rArr (dy)/(dx)=x^(x)(1+logx).` |
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| 339. |
Prove that `(d)/(dx)(cos^(-1)x)=(1)/(sqrt(1-x^(2))`, where `x in [-1,1].` |
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Answer» Let `y=cos^(-1)x`, where `x in [-1,1] and y- in [-(pi)/(2),(pi)/(2)].` Then, `y=cos^(-1)x rArr x= cos y` `rArr(dy)/(dx)=-siny," where sin "ygt0," since y"in[0,(pi)/(2)]` `rArr(dy)/(dx)=-sqrt(1-cos^(2)y)=-sqrt(1-x^(2))` `rArr(dy)/(dx)=(-1)/(sqrt(1-x^(2)))` Hence, `(d)/(dx)(cos^(-1)x)=(-1)/(sqrt(1-x^(2)))`. |
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| 340. |
The value of `(d)/(dx) cot x^(@)` is -A. `-"cosec"^(2) x^(@)`B. `"cosec"^(2)x^(@)`C. `-(180)/(pi)"cosec"^(2)x^(@)`D. `-(pi)/(180) "cosec"^(2)x^(@)` |
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Answer» Correct Answer - D |
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| 341. |
Find `(dy)/(dx)`for the function:`y=sin^(-1)sqrt((1-x))+cos^(-1)sqrt(x)` |
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Answer» Correct Answer - `(-1)/(sqrt(x-x^(2)))` `sin^(-1)sqrt(1-x)=sin^(-1)sqrt(1(sqrt(x))^(2))=cos^(-1)sqrt(x)` `therefore" "y=2 cos^(-1)sqrt(x)` `"or "(dy)/(dx)=2xx(-1)/(sqrt(1-x))xx(1)/(2sqrt(xx))=(-1)/(sqrt(1-x^(2))` |
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| 342. |
`y=e^(sin x^(3))` |
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Answer» Correct Answer - `e^(sin x^(2))cos x^(2)2x` `"Let "y=e^(sinx^(2)).` Putting `x^(2)=v and u=sin x^(2) = sin v,` we get `y=e^(u),u=sin v, and v=x^(2)` `therefore" "(dy)/(du)=e^(u),(du)/(dv)=cos v, and (dv)/(dx)=2x` `"Now, "(dy)/(dx)xx(dy)/(du)xx(du)/(dv)xx(dv)/(dx)` `=e^(u)cos v 2=e^(sin v)cos v 2x` `=e^(sin x^(2))cos x^(2)2x` |
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| 343. |
`"If "y=(tan x)^((tan x)^(tan x))," then find "(dy)/(dx).` |
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Answer» Correct Answer - `y log y sec^(2)x[log tan x + 1 +(1)/(tan x log tan x)]` Taking logarithm on both sides, we get `log y = (tan x)^(tan x)log tan x` `therefore" "log log y = [ tan x log tan x ] + log log tan x ` Differentiating w.r.t x, we get `(1)/(y log y )(dy)/(dx)=sec^(2) x log tan x + tan x (sec^(2)x)/(tan x)+(1)/(log tan x )xx(sec^(2)x)/(tan x)` `therefore" "(dy)/(dx)=y log y sec^(2) x [ log tan x + 1+ //(tan x log tan x )]` |
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| 344. |
`"If "x=(2costheta-cos 2theta)and y=(2sin theta-sin 2theta)," find "((d^(2))/(dx^(2)))_(theta=(pi)/(2))` |
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Answer» Correct Answer - `(-3)/(2)` `(dy)/(dx)=((dy//d theta))/((dx//d theta))=tan.(3 theta)/(2)` `rArr(d^(2)y)/(dx^(2))=(3)/(2)sec^(2)(3theta)/(2).(d theta)/(dx)=(3)/(2)sec^(2).(3theta)/(2).(1)/(2(sin 2theta-sin theta))` `rArr((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(3)/(4)sec^(2).(3pi)/(4).(1)/((sin pi-sin.(pi)/(2)))=(3)/(4)xx2xx(1)/((0-1))=(-3)/(2).` |
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| 345. |
Prove that `(d)/(dx)(cot^(-1)x)=(-1)/((1+x^(2)))`, where `x in R`. |
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Answer» Let `y=cot(-1)x`, where `x in R and y in [0,pi]`. Then, `x=cot y` `rArr(dx)/(dy)=-"cosec"^(2)y=-(1+cot^(2)y)=-(1+x^(2))` `rArr(dy)/(dx)=(-1)/((1+x^(2))).` ltbr. Hence, `(d)/(dx)(cot^(-1)x)=(-1)/((1+x^(2))).` |
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| 346. |
Differentiate `(sin x)^(x)` w.r.t. x. |
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Answer» Let `y=(sin x)^(x)." …(i)" ` Taking logarithm on both sides of (i), we get `log y= x log (sinx)." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get |
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| 347. |
If `y=sin^(-1){(sqrt(1+x)-sqrt(1-x))/(2)}`, find `(dy)/(dx).` |
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Answer» Putting `x=cos 2 theta,` we get `y=sin^(-1){(sqrt(1+cos 2theta)-sqrt(1-cos2 theta))/(2)}` `=sin^(-1){(sqrt(2cos^(2)theta)-sqrt(2sin^(2)theta))/(2)}=sin^(-1){(sqrt(2)cos^(2)theta-sqrt(2sin^(2)theta))/(2)}=sin^(-1){(sqrt2cos theta-sqrt2 sin theta)/(2)}` `=sin^(-1){(1)/(sqrt2)cos theta-(1)/(sqrt2)sin theta}=sin^(-1){"sin"(pi)/(4)cos theta-"cos"(pi)/(4)sin theta}` `=sin^(-1){sin((pi)/(4)-theta)}` `=((pi)/(4)-theta)=((pi)/(4)-(1)/(2)cos^(-1)x)` `" "[becausex=cos 2 thetarArr 2 theta=cos^(-1)x rArr theta=(1)/(2)cos^(-1)x]` `thereforey=(pi)/(4)-(1)/(2)cos^(-1)x.` Hence, `(dy)/(dx)=(d)/(dx){(pi)/(4)-(1)/(2)cos^(-1)x}=(d)/(dx)((pi)/(4))-(1)/(2)(d)/(dx)(cos^(-1)x)` `={0-(1)/(2).((-1))/(sqrt(1-x^(2)))}=(1)/(2sqrt(1-x^(2))).` |
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| 348. |
Differentiate `tan^(-1){(x^(1//3)+a^(1//3))/(1-(a x)^(1//3))}`with respect to `x` |
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Answer» Let `y=((x^(1//3)+a^(1//3))/(1-x^(1//3)a^(1//3))).` Putting `x^(1//3)=tan theta and a^(1//3)=tan phi,` we get `y=tan^(-1)((tan theta+tan phi)/(1-tan theta tan phi))=tan^(-1)[tan(theta+phi)]` `=(theta+phi)=tan^(-1)(a^(1//3)).` `therefore(dy)/(dx)=(d)/(dx){tan^(-1)(x^(1//3))+tan^(-1)(a^(1//3))}` `=(d)/(dx){tan^(-1)(x^(1//3))+tan^(-1)(a^(1//3))}.` `=(d)/(dx){tan^(-1)(x^(1//3))}+(d)/(dx){tan^(-1)(a^(1//3))}=(1)/((1+x^(2//3))).(1)/(3)x^(-2//3)` `=(1)/(3x^(2//3)(1+x^(2//3)))" "[because" "tan^(-1)(a^(1//3))="constant"].` |
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| 349. |
Find the derivative of `tan^(-1)(2x)/(1-x^2)wdotrdottsin^(-1)(2x)/(1+x^2)` |
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Answer» Correct Answer - 1 `"Let "y=tan^(-1)""(2x)/(1-x^(2))=2tan^(-1)x` `"and "z= sin^(-1)""(2x)/(1+x^(2))=2tan^(-1)x` `therefore" "(dy)/(dx)=1` |
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| 350. |
If `x= 2costheta-cos 2theta` and `y=2 sintheta - sin 2theta`.Find `(d^2y)/(dx^2)` at `theta=pi/2` |
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Answer» We have `x=(2cos theta-cos 2 theta)and y=(2 sin theta-sin 2 theta)` `rArr(dx)/(d theta)=(-2sin theta+2 sin 2 theta)and (dy)/(d theta)=(2cos theta-2 cos 2 theta)` `rArr(dy)/(dx)=((dy//d theta))/((dx//d theta))=((2cos theta-2 cos 2theta))/((-2sin theta+2sin 2 theta))=((cos theta-cos 2 theta))/((sin 2 theta-sin theta))` `=(2sin((3 theta)/(2))sin.(theta)/(2))/(2cos((3theta)/(2))sin.(theta)/(2))=tan.(3theta)/(2)` `rArr(d^(2)y)/(dx^(2))=(d)/(dx)(tan.(3theta)/(2))=(3)/(2)sec^(2).(3theta)/(2).(d theta)/(dx)=(3)/(2)sec^(2).(3 theta)/(2).(1)/(2(sin 2 theta-sin theta))` `rArr((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(3)/(2).sec^(2)((3pi)/(4)).(1)/((2sinpi-sin.(pi)/(2)))` `=(-3)/(4)sec^(2).(pi)/(4)=(-3)/(4)xx(sqrt2)^(2)=(-3)/(2)` `" "[becausesec.(3pi)/(4)=sec(pi-(pi)/(4))=-sec.(pi)/(4)].` |
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