80 + Interview Questions in DERIVATIVES IN GENERAL AWARENESS Page 2 InterviewSolution

51.	Let `f(x)=[\|x\|-\|x-1\|]^(2)` Which of the following equation is/are correct? 1. `f(-2)=f(5)` 2. `f'(-2)+f'(0.5)+f'(3)=4` Select the correct answer using the code given below:A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2
Answer» Correct Answer - A Given `f(x)=(\|x\|-\|x+1\|)^(2)` `f(x)={{:(1" "x le 0),((2x-1)^(2)" "0 lt x lt 1),(1" "x ge 1):}` For x = -2 f(x) = 1 so f(-2) = 1 For x = 5 f(x) = 1 `rArr` f(5) = 1 Hence f(-2) = f(5) Now, for x = -1 f'(x) = 0 f'(-2) = 0 For x = 0.5 `f'(x) = 8 rArr f'(0.5) = 8` For x = 3 `f'(x) = 0 rArr f'(3) = 0` `rArr f'(-2) + f'(0.5) + f'(3) = 8 ne 4`

Discussion

52.	y = (ax+b)m
Answer» The required condition is *n=m+1*

52.

y = (ax+b)m

Answer»

The required condition is

n=m+1

Discussion

53.	If y = a sin mx + b cos mx, then \(\frac{d^2y}{dx^2}\) is equal to A. –m2y B. m2y C. –my D. my
Answer» Correct Answer is (A) –m²y Given: y = a sin mx + b cos mx \(\frac{dy}{dx}\)= ma cos mx - mb sin mx \(\frac{d^2y}{dx^2}\)= -m² a sin mx - m² b cos mx = -m² [a sin mx + b cos mx] = - m²y

53.

If y = a sin mx + b cos mx, then \(\frac{d^2y}{dx^2}\) is equal to A. –m2y B. m2y C. –my D. my

Answer»

Correct Answer is (A) –m²y

Given:

y = a sin mx + b cos mx

\(\frac{dy}{dx}\)= ma cos mx - mb sin mx

\(\frac{d^2y}{dx^2}\)= -m² a sin mx - m² b cos mx

= -m² [a sin mx + b cos mx]

= - m²y

Discussion

54.	If `y=ln(e^(mx)+e^(-mx))`, then what is `(dy)/(dx)` at x = 0 equal to ?A. `-1`B. 0C. 1D. 2
Answer» Correct Answer - B `y=ln(emx+e-mx)` `Phi" "(dy)/(dx)=(1)/(e^(mx)+e^(-mx)).(d)/(dx)(emx+e-mx)` `=(me^(mx)-me^(-mx))/(e^(mn)+e^(-mx))=(m(e^(mx)-e^(-mx)))/(e^(mx)+e^(-mx))` `=(m[e^(mn)-(1)/(e^(mx))])/(e^(mx)+(1)/(x^(nx)))=(m[e^(2mx)-1])/(e^(2mx)+1)` so, `(dy)/(dx):\|_(x=0)=(m(e_(0)-1))/(e^(0)+1)=m(0)=0`

Discussion

55.	Find the derivative of \(\frac{1}{x^{11}}.\)
Answer» Let, \(y=\frac{1}{x^{11}}\) \(=x^{-11}\) ∴ \(\frac{dy}{dx}=\frac{d}{dx}(x^{-11})\) \(=-11\,x^{-11-1}\) \(=-\frac{11}{x^{12}}\)

55.

Find the derivative of \(\frac{1}{x^{11}}.\)

Answer»

Let,

\(y=\frac{1}{x^{11}}\)

\(=x^{-11}\)

∴ \(\frac{dy}{dx}=\frac{d}{dx}(x^{-11})\)

\(=-11\,x^{-11-1}\)

\(=-\frac{11}{x^{12}}\)

Discussion

56.	If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=2xy\)B. \(x\frac{d^2y}{dx^2}=y_1\)C. \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)D.\(x\frac{d^2y}{dx^2}=2xy\)
Answer» Correct Answer is (C) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\) Given: \(\frac{dy}{dx}=2bx\) \(\frac{d^2y}{dx^2}=2b\neq2xy\) \(x\frac{d^2y}{dx^2}=2bx\) \(=\frac{dy}{dx}\) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y\) \(=2bx-2bx+y\) = y

56.

If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=2xy\)B. \(x\frac{d^2y}{dx^2}=y_1\)C. \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)D.\(x\frac{d^2y}{dx^2}=2xy\)

Answer»

Correct Answer is (C) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)

Given:

\(\frac{dy}{dx}=2bx\)

\(\frac{d^2y}{dx^2}=2b\neq2xy\)

\(x\frac{d^2y}{dx^2}=2bx\)

\(=\frac{dy}{dx}\)

\(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y\) \(=2bx-2bx+y\)

= y

Discussion

57.	Find the derivative of (x2 + 1) cos x.
Answer» Let, y = (x² + 1) cos x ∴ \(\frac{dy}{dx}=\frac{d}{dx}[(x^2+1)cos\,x]\) \(=cos\,x\frac{d}{dx}(x^2+1)+(x^2+1)\frac{d}{dx}cos\,x\) \(=cos\,x.2x+(x^2+1).(-sin\,x)\) = 2x cos x − x² sin x − sin x

57.

Find the derivative of (x2 + 1) cos x.

Answer»

Let,

y = (x² + 1) cos x

∴ \(\frac{dy}{dx}=\frac{d}{dx}[(x^2+1)cos\,x]\)

\(=cos\,x\frac{d}{dx}(x^2+1)+(x^2+1)\frac{d}{dx}cos\,x\)

\(=cos\,x.2x+(x^2+1).(-sin\,x)\)

= 2x cos x − x² sin x − sin x

Discussion

58.	If `y=sin(ax+b)`, then what is `(d^(2)y)/(dx^(2))` at `x=-(b)/(a)`, where a, be are constants and `a ne 0`?
Answer» Correct Answer - A Let y = sin(ax + b) `rArr (dy)/(dx) = a cos(ax + b)` `rArr (d^(2)y)/(dx^(2))= -a^(2) sin(ax+b)` Now, `(d^(2)y)/(dx^(2)) "at x" = -(b)/(a)` is `-a^(2)sin(a(-(b)/(a))+b)=-a^(2) sin 0 = 0`

Discussion

59.	Find the derivative of x2 - 2 at x = 10 from first principle.
Answer» Let, f(x) = x² − 2 ∴ f' (x) = \(\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) = \(\lim\limits_{h \to 0}\frac{[(x+h)^2-2]-(x^2-2)}{h}\) = \(\lim\limits_{h \to 0}\frac{x^2+2xh+h^2-2-x^2+2}{h}\) = \(\lim\limits_{h \to 0}\frac{2xh+h^2}{h}\) = \(\lim\limits_{h \to 0}(2x+h)\) = 2x + 0 = 2x ∴ At x = 10, f'(x) = 2 ∙10 = 20.

59.

Find the derivative of x2 - 2 at x = 10 from first principle.

Answer»

Let,

f(x) = x² − 2

∴ f' (x) = \(\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

= \(\lim\limits_{h \to 0}\frac{[(x+h)^2-2]-(x^2-2)}{h}\)

= \(\lim\limits_{h \to 0}\frac{x^2+2xh+h^2-2-x^2+2}{h}\)

= \(\lim\limits_{h \to 0}\frac{2xh+h^2}{h}\)

= \(\lim\limits_{h \to 0}(2x+h)\)

= 2x + 0

= 2x

∴ At x = 10,

f'(x) = 2 ∙10

= 20.

Discussion

60.	Find the derivative of \(\sqrt{4-x}\) from first principle.
Answer» Let, \(f(x)=\sqrt{4-x}\) ∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) \(=\lim\limits_{h \to 0}\frac{\sqrt{4-(x+h)}-\sqrt{4-x}}{h}\) \(=\lim\limits_{h \to 0}\frac{[\sqrt{4-(x+h)}-\sqrt{4-x}][\sqrt{4-(x+h)}+\sqrt{4-x}]}{[h\sqrt{4-(x+h)}+\sqrt{4-x}]}\) \(=\lim\limits_{h \to 0}\frac{[{4-(x+h)}]-(4-x)}{h[\sqrt{4-(x+h)}+\sqrt{4-x}]}\) \(=\lim\limits_{h \to 0}\frac{-h}{h\sqrt{4-(x+h)}+\sqrt{4-x}}\) \(=\lim\limits_{h \to 0}\frac{1}{\sqrt{4-(x+h)}+\sqrt{4-x}}\) = \(-\frac{1}{2\sqrt{4-x}}\)

60.

Find the derivative of \(\sqrt{4-x}\) from first principle.

Answer»

Let,

\(f(x)=\sqrt{4-x}\)

∴ \(f'(x)=\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\(=\lim\limits_{h \to 0}\frac{\sqrt{4-(x+h)}-\sqrt{4-x}}{h}\)

\(=\lim\limits_{h \to 0}\frac{[\sqrt{4-(x+h)}-\sqrt{4-x}][\sqrt{4-(x+h)}+\sqrt{4-x}]}{[h\sqrt{4-(x+h)}+\sqrt{4-x}]}\)

\(=\lim\limits_{h \to 0}\frac{[{4-(x+h)}]-(4-x)}{h[\sqrt{4-(x+h)}+\sqrt{4-x}]}\)

\(=\lim\limits_{h \to 0}\frac{-h}{h\sqrt{4-(x+h)}+\sqrt{4-x}}\)

\(=\lim\limits_{h \to 0}\frac{1}{\sqrt{4-(x+h)}+\sqrt{4-x}}\)

= \(-\frac{1}{2\sqrt{4-x}}\)

Discussion

61.	Find the derivative of cot3x.
Answer» Let, y = cot³x ∴ \(\frac{dy}{dx}=\frac{d}{dx}(cot^3\,x)\) \(=3\,cot^2\frac{d}{dx}(cot\,x)\) \(=3\,cot^2\,x(-cosec^2\,x)\) \(=3\,cot^2\,x.cosec^2\,x\)

61.

Find the derivative of cot3x.

Answer»

Let,

y = cot³x

∴ \(\frac{dy}{dx}=\frac{d}{dx}(cot^3\,x)\)

\(=3\,cot^2\frac{d}{dx}(cot\,x)\)

\(=3\,cot^2\,x(-cosec^2\,x)\)

\(=3\,cot^2\,x.cosec^2\,x\)

Discussion

62.	If `y=sin^(-1)(x-y),x=3t,y=4t^(3)`, then what is the derivative of u with respect to t?A. `3(1-t^(2))`B. `3(1-t^(2))^(-(1)/(2))`C. `5(1-t^(2))^((1)/(2))`D. `5(1-t^(2))`
Answer» Correct Answer - B `u=sin^(-1)(x-y)x=3t,y=4t^(3)` So, `u=sin^(-1)(3t-4t^(3))` Let `t=sin theta rArr theta=sin^(-1)t`, So, `u=sin^(-1)(3sin theta-4sin^(3)theta)` `=sin^(-1)(sin3theta)=3 theta` Hence, `u=3sin^(-1)t` `(du)/(dt)=3(1)/(sqrt(1-t^(2)))=3(1-t^(2))^(-1//2)`

Discussion

63.	If `y=x+e^x ,`then `(d^2x)/(dy^2)`is equal toA. `e^(x)`B. `-(e^(x))/((1+e^(x))^(3))`C. `-(e^(x))/((1+e^(x))`D. `(e^(x))/((1+e^(x))^(2))`
Answer» Correct Answer - B Givne that `y=x+e^(x)` Differentiating w.r.t. x `(dy)/(dx)=1+e^(x)` `rArr" "(dx)/(dy)=(1)/(1+e^(x))` Differentiating w.r.t. y `(d^(2)x)/(dy^(2))=-((1)(e^(x)))/((1+e^(x))^(2)).(dx)/(dy)` `=-(e^(x))/((1+e^(x))^(2)).(1)/((1+e^(x)))=-(e^(x))/((1+e^(x))^(3))`

Discussion

64.	If `f(x)=(x-2)/(x+2), x ne 2,` then what is `f^(-1)(x)` equal to?A. `(4(x+2))/(x-2)`B. `(x+2)/(4(x-2))`C. `(x+2)/(x-2)`D. `(2(1+x))/(1-x)`
Answer» Correct Answer - D

Discussion

65.	If `y=sin^(-1)x+sin^(-1)sqrt(1-x^(2))`, what is `(dy)/(dx)` equal to?A. `cos^(-1)x+cos^(-1)sqrt(1-x^(2))`B. `(1)/(cosx)+(1)/(cos sqrt(1-x^(2)))`C. `(pi)/(2)`D. 0
Answer» Correct Answer - D Given function is : `y=sin^(-1)x+sin^(-1)sqrt(1-x^(2))` On differentiating, w.r.t. x, we get `(dy)/(dx)=(1)/(sqrt(1-x^(2)))+(1)/(sqrt(1-1+x^(2))).(1)/(2sqrt(1-x^(2)))(-2x)` `=(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-x^(2)))=0`

Discussion

66.	Find the first principle the derivative of sin2x.
Answer» Let, f(x) = sin²x ∴ \(f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) \( = \lim\limits_{h \to 0}\frac{sin^2(x+h)-sin^2x}{h}\) \( = \lim\limits_{h \to 0}\frac{\{sin(x+h)+sinx\}\{sin(x+h)-sin\,x\}}{h}\) \( = \lim\limits_{h \to 0}\frac{(2sin\frac{x+h+x}{2}cos\frac{x+h-x}{2}).(2cos\frac{x+h+x}{2}sin\frac{x+h-x}{2})}{h}\) \( = \lim\limits_{h \to 0}\frac{4cos\frac{h}{2}sin\frac{h}{2}.cos(x+\frac{h}{2})sin(x+\frac{h}{2})}{h}\) \( = 2\lim\limits_{h \to 0}\frac {sin \frac{h}{2}}{\frac{h}{2}}\)\( \lim\limits_{h \to 0}\cos\frac{h}{2}cos(x+\frac{h}{2})\) = 2 ∙ 1 ∙ 1 ∙ cos x ∙ sin x = 2 cos x sin x = sin 2x.

66.

Find the first principle the derivative of sin2x.

Answer»

Let,

f(x) = sin²x

∴ \(f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\( = \lim\limits_{h \to 0}\frac{sin^2(x+h)-sin^2x}{h}\)

\( = \lim\limits_{h \to 0}\frac{\{sin(x+h)+sinx\}\{sin(x+h)-sin\,x\}}{h}\)

\( = \lim\limits_{h \to 0}\frac{(2sin\frac{x+h+x}{2}cos\frac{x+h-x}{2}).(2cos\frac{x+h+x}{2}sin\frac{x+h-x}{2})}{h}\)

\( = \lim\limits_{h \to 0}\frac{4cos\frac{h}{2}sin\frac{h}{2}.cos(x+\frac{h}{2})sin(x+\frac{h}{2})}{h}\)

\( = 2\lim\limits_{h \to 0}\frac {sin \frac{h}{2}}{\frac{h}{2}}\)\( \lim\limits_{h \to 0}\cos\frac{h}{2}cos(x+\frac{h}{2})\)

= 2 ∙ 1 ∙ 1 ∙ cos x ∙ sin x

= 2 cos x sin x

= sin 2x.

Discussion

67.	If `x=cost, y=sint,` then what is `(d^(2)y)/(dx^(2))` equal to?A. `y^(-3)`B. `y^(3)`C. `-y^(-3)`D. `-y^(3)`
Answer» Correct Answer - C Given that `x=cost, y=sint` `rArr" "(dx)/(dt)-sint and (dy)/(dt)=cot` `rArr" "(dy)/(dx)=((dy)/(dt))/((dx)/(dt))=-(cost)/(sint)rArr (dy)/(dx)=-cot t` `rArr" "(d^(2)y)/(dx^(2))="cosec"^(2)t.(dt)/(dx)="cosec"^(2)t.(1)/(-sint)=(1)/(sin^(3)t)` `rArr" "(d^(2)y)/(dx^(2))=-y^(-3)`

Discussion

68.	If `y=tan^(-1)((5-2tansqrtx)/(2+5tansqrtx))`, then what is `(dy)/(dx)` equal to ?A. `-(1)/(2sqrtx)`B. 1C. `-1`D. `(1)/(2sqrtx)`
Answer» Correct Answer - A

Discussion

69.	If `y=e^(x^(2))sin2x`, then what is `(dy)/(dx)` at `x=pi` equal to?A. `(1+pi)e^(pi^(2))`B. `2pie^(pi^(2))`C. `2e^(pi^(2))`D. `e^(pi^(2))`
Answer» Correct Answer - C

Discussion

70.	For the curve `sqrtx+sqrty=1`, what is the value of `(dy)/(dx)` at `((1)/(4),(1)/(4))`?A. `(1)/(2)`B. 1C. `-1`D. 2
Answer» Correct Answer - C Given function : `sqrtx+sqrty=1` is an implicit functionb Differentiating both sides w.r.t. x, we get `(1)/(2sqrtx)+(1)/(2sqrty)(dy)/(dx)=0` `rArr(dy)/(dx)=-sqrt((y)/(x))` Value of `(dy)/(dx)" at "x=(1)/(4),y=(1)/(4)` `((dy)/(dx))_(((1)/(4)","(1)/(4)))=-sqrt((1//4)/(1//4))=-1`

Discussion

71.	What is the derivative of `tan^(-1)((sqrtx-x)/(1+x^(3//2)))` at x = 1?A. `-(1)/(4)`B. `(1)/(2)`C. `(3)/(2)`D. 1
Answer» Correct Answer - A Let `y=tan^(-1)((sqrtx-x)/(1+x^(3//2)))=tan^(-1).(sqrtx-x)/(1+sqrtx.x)` `=tan^(-1)sqrtx-tan^(-1)x` On differentiating w.r.t. we get `(dy)/(dx)=(1)/(1+x).(1)/(2sqrtx)-(1)/(1+x^(2))` Now, `((dy)/(dx))_(x=1)=(1)/(1+1).(1)/(2)-(1)/(1+1)=(1)/(4)-(1)/(2)=-(1)/(4).`

Discussion

72.	Find the derivative of the following function:\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)
Answer» Let, \(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\) \((sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)\) \(∴f'(x)=\frac{(sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)\frac{d}{dx}(sin\,x-cos\,x)}{(sin\,x-cos\,x)^2}\) \(=\frac{-(sin\,x-cos\,x)(cos\,x-sin\,x)-(sin\,x+cos\,x)(sin\,x+cos\,x)}{(sin\,x-cos\,x)^2}\) \(=\frac{-sin^2\,x+2sin\,x\,cos\,x\,-cos^2\,x\,-sin^2\,x-2\,sin\,x\,cos\,x-cos^2\,x}{(sin\,x\,-cos\,x)^2}\)\(=\frac{-2\,(sin^x+cos^2x)}{(sin\,x-cos^\,x)^2}\) \(=\frac{-2}{(sin\,x-cos^\,x)^2}\) \(f(x)={(sinx+cosx)\over(sinx−cosx)}\) =>\(f(x)=-{(1+tanx)\over(1-tanx)}\) => \(f(x)=-{tan(pi/4+x)}\) =>\(f'(x)=-{sec^2(pi/4+x)}\)

72.

Find the derivative of the following function:\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)

Answer»

Let,

\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)

\((sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)\)

\(∴f'(x)=\frac{(sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)\frac{d}{dx}(sin\,x-cos\,x)}{(sin\,x-cos\,x)^2}\) \(=\frac{-(sin\,x-cos\,x)(cos\,x-sin\,x)-(sin\,x+cos\,x)(sin\,x+cos\,x)}{(sin\,x-cos\,x)^2}\) \(=\frac{-sin^2\,x+2sin\,x\,cos\,x\,-cos^2\,x\,-sin^2\,x-2\,sin\,x\,cos\,x-cos^2\,x}{(sin\,x\,-cos\,x)^2}\)\(=\frac{-2\,(sin^x+cos^2x)}{(sin\,x-cos^\,x)^2}\)

\(=\frac{-2}{(sin\,x-cos^\,x)^2}\)

\(f(x)={(sinx+cosx)\over(sinx−cosx)}\)

=>\(f(x)=-{(1+tanx)\over(1-tanx)}\)

=> \(f(x)=-{tan(pi/4+x)}\)

=>\(f'(x)=-{sec^2(pi/4+x)}\)

Discussion

73.	If `y=cost and x=sint`, then what is `(dy)/(dx)` equal to?A. xyB. `x//y`C. `-y//x`D. `-x//y`
Answer» Correct Answer - D Let `y=cost, x=sint` `(dy)/(dt)=-sint,(dx)/(dt)=cost` `(dy)/(dx)=(dy//dt)/(dx//dt)=-(sint)/(cost)=-(x)/(y)`

Discussion

74.	What is the derivative of `sqrt((1+cosx)/(1-cosx))`?A. `(1)/(2)sec^(2).(x)/(2)`B. `-(1)/(2)"cosec"^(2_.(x)/(2)`C. `-"cosec"^(2).(x)/(2)`D. None of these
Answer» Correct Answer - B Let `y = sqrt((1+cos x)/(1-cos x))=(sqrt(2)"cos"(x)/(2))/(sqrt(2)"sin"(x)/(2))="cot"(x)/(2)` `(dy)/(dx)=-"cosec"^(2)(x)/(2).(1)/(2)=-(1)/(2)"cosec"^(2)(x)/(1)`

Discussion

75.	If `x = t^(2)` and `y = t^(3)`, then `(d^(2)y)/(dx^(2))` is equal toA. 1B. `(3)/(2t)`C. `(3)/(4t)`D. `(3)/(2)`
Answer» Correct Answer - C Let `x=t^(2) and y=t^(3)` `rArr" "(dx)/(dt=2t and (dy)/(dt)=3t^(2)` `therefore" "(dy)/(dx)=(dy//dt)/(dx//dt)=(3t^(2))/(2t)=(3)/(2)t` `rArr" "(d^(2)y)/(dx^(2))=(3)/(2).(dt)/(dx)=(3)/(2).(1)/(2t)(because(dx)/(dt)=2t)` `=(3)/(4t)`

Discussion

76.	If `sqrtx+sqrty=2`, then what is `(dy)/(dx)` at y = 1 and x = 1 equal to ?A. 5B. 2C. 4D. `-1`
Answer» Correct Answer - D Let `sqrtx+sqrty=2` Differentiate w.r.t. x, we get `(1)/(2sqrtx)+(1)/(2sqrty)(dy)/(dx)=0" ...(1)"` Put `y=1, x=1` in equation (1) `(1)/(2)+(1)/(2)(dy)/(dx)=0` `rArr" "(dy)/(dx)=-1`

Discussion

77.	What is the derivative of `sin(sinx)?`A. `cos(cosx)`B. `cos(sinx)`C. `cos(sinx)cosx`D. `cos(cosx)cosx`
Answer» Correct Answer - C `(d)/(dx) sin(sin x) = cos(sin x). "cos x"`

Discussion

78.	What is the derivative of `x^(3)` with respect to `x^(2)`?A. `3x^(2)`B. `(3x)/(2)`C. `x`D. `(3)/(2)`
Answer» Correct Answer - B `U = x^(3)` `(dU)/(dx)=3x^(2)" "...(1)` `V = x^(2)` `(dV)/(dx)=2x` From (1) and (2) `(dU)/(dV)=(3x^(2))/(2x)=(3)/(2)x`

Discussion

79.	What is the derivative of `log_(x)5` with respect to `log_(5)x`?A. `-(log_(5)x)^(-2)`B. `(log_(5)x)^(-2)`C. `-(log_(x)5)^(-2)`D. `(log_(x)5)^(-2)`
Answer» Correct Answer - A Let `u_(1)=log_(x)5 and u_(2)=log_(5)x` `rArr u_(1)=(log_(e)5)/(log_(e)x) and u_(2)=(log_(e)x)/(log_(e)5)` On differentiating w.r.t. x, we get `(du_(1))/(dx)[(log_(e)x(0)-((1)/(x)))/((log_(e)x)^(2))]log_(e)5=-(log_(e)5)/(x(log_(e)x))` `and (du_(2))/(dx)=(1)/(xlog_(e)5)` `therefore(du_(1))/(du_(2))=(du_(1)//dx)/(du_(2)//dx)=-(log_(e)5)/(x(log_(e)x)^(2))xx x log_(e)5` `=-((log_(e)5)/(log_(e)x))^(2)=-(log_(x)5)^(2)=-(log_(5)x)^(-2)`

Discussion

80.	Consider the curve `x=a(cos theta+thetasin theta)and y=a(sin theta-thetacos theta).` If `y=xln x+xe^(x)`, then what is the value of `(dy)/(dx)` at x = 1?A. `1+e`B. `1-e`C. `1+2e`D. None of these
Answer» Correct Answer - C `y = "x In x" + xe^(x)` After differentiating both sides with respect to x, we get `(dy)/(dx)=x.(1)/(x)+log x + xe^(x) + e^(x)` or `=1 + log x + xe^(x) + e^(x)` Therefore `((dy)/(dx))_(x=1)=1 + log 1 + 1.e^(1) + e^(1) = 1 = 2e" "[because log 1 = 0]`

Discussion

Explore topic-wise InterviewSolutions in .

Let `f(x)=[|x|-|x-1|]^(2)` Which of the following equation is/are correct? 1. `f(-2)=f(5)` 2. `f'(-2)+f'(0.5)+f'(3)=4` Select the correct answer using the code given below:A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2

y = (ax+b)m

If y = a sin mx + b cos mx, then \(\frac{d^2y}{dx^2}\) is equal to A. –m2y B. m2y C. –my D. my

If `y=ln(e^(mx)+e^(-mx))`, then what is `(dy)/(dx)` at x = 0 equal to ?A. `-1`B. 0C. 1D. 2

Find the derivative of \(\frac{1}{x^{11}}.\)

If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=2xy\)B. \(x\frac{d^2y}{dx^2}=y_1\)C. \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)D.\(x\frac{d^2y}{dx^2}=2xy\)

Find the derivative of (x2 + 1) cos x.

If `y=sin(ax+b)`, then what is `(d^(2)y)/(dx^(2))` at `x=-(b)/(a)`, where a, be are constants and `a ne 0`?

Find the derivative of x2 - 2 at x = 10 from first principle.

Find the derivative of \(\sqrt{4-x}\) from first principle.

Find the derivative of cot3x.

If `y=sin^(-1)(x-y),x=3t,y=4t^(3)`, then what is the derivative of u with respect to t?A. `3(1-t^(2))`B. `3(1-t^(2))^(-(1)/(2))`C. `5(1-t^(2))^((1)/(2))`D. `5(1-t^(2))`

If `y=x+e^x ,`then `(d^2x)/(dy^2)`is equal toA. `e^(x)`B. `-(e^(x))/((1+e^(x))^(3))`C. `-(e^(x))/((1+e^(x))`D. `(e^(x))/((1+e^(x))^(2))`

If `f(x)=(x-2)/(x+2), x ne 2,` then what is `f^(-1)(x)` equal to?A. `(4(x+2))/(x-2)`B. `(x+2)/(4(x-2))`C. `(x+2)/(x-2)`D. `(2(1+x))/(1-x)`

If `y=sin^(-1)x+sin^(-1)sqrt(1-x^(2))`, what is `(dy)/(dx)` equal to?A. `cos^(-1)x+cos^(-1)sqrt(1-x^(2))`B. `(1)/(cosx)+(1)/(cos sqrt(1-x^(2)))`C. `(pi)/(2)`D. 0

Find the first principle the derivative of sin2x.

If `x=cost, y=sint,` then what is `(d^(2)y)/(dx^(2))` equal to?A. `y^(-3)`B. `y^(3)`C. `-y^(-3)`D. `-y^(3)`

If `y=tan^(-1)((5-2tansqrtx)/(2+5tansqrtx))`, then what is `(dy)/(dx)` equal to ?A. `-(1)/(2sqrtx)`B. 1C. `-1`D. `(1)/(2sqrtx)`

If `y=e^(x^(2))sin2x`, then what is `(dy)/(dx)` at `x=pi` equal to?A. `(1+pi)e^(pi^(2))`B. `2pie^(pi^(2))`C. `2e^(pi^(2))`D. `e^(pi^(2))`

For the curve `sqrtx+sqrty=1`, what is the value of `(dy)/(dx)` at `((1)/(4),(1)/(4))`?A. `(1)/(2)`B. 1C. `-1`D. 2

What is the derivative of `tan^(-1)((sqrtx-x)/(1+x^(3//2)))` at x = 1?A. `-(1)/(4)`B. `(1)/(2)`C. `(3)/(2)`D. 1

Find the derivative of the following function:\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)

If `y=cost and x=sint`, then what is `(dy)/(dx)` equal to?A. xyB. `x//y`C. `-y//x`D. `-x//y`

What is the derivative of `sqrt((1+cosx)/(1-cosx))`?A. `(1)/(2)sec^(2).(x)/(2)`B. `-(1)/(2)"cosec"^(2_.(x)/(2)`C. `-"cosec"^(2).(x)/(2)`D. None of these

If `x = t^(2)` and `y = t^(3)`, then `(d^(2)y)/(dx^(2))` is equal toA. 1B. `(3)/(2t)`C. `(3)/(4t)`D. `(3)/(2)`

If `sqrtx+sqrty=2`, then what is `(dy)/(dx)` at y = 1 and x = 1 equal to ?A. 5B. 2C. 4D. `-1`

What is the derivative of `sin(sinx)?`A. `cos(cosx)`B. `cos(sinx)`C. `cos(sinx)cosx`D. `cos(cosx)cosx`

What is the derivative of `x^(3)` with respect to `x^(2)`?A. `3x^(2)`B. `(3x)/(2)`C. `x`D. `(3)/(2)`

What is the derivative of `log_(x)5` with respect to `log_(5)x`?A. `-(log_(5)x)^(-2)`B. `(log_(5)x)^(-2)`C. `-(log_(x)5)^(-2)`D. `(log_(x)5)^(-2)`

Consider the curve `x=a(cos theta+thetasin theta)and y=a(sin theta-thetacos theta).` If `y=xln x+xe^(x)`, then what is the value of `(dy)/(dx)` at x = 1?A. `1+e`B. `1-e`C. `1+2e`D. None of these