If $$\sin ^{ 2 }{ mx } +\cos ^{ 2 }{ ny } ={ a }^{ 2 }$$ then $$\cfrac{dy}{dx}=$$

$$\cfrac{m.\sin{2mx}}{n.\sin{2ny}}$$

$$\cfrac{m.\sin{mx}}{n.\sin{nx}}$$

$$\cfrac{-m.\cos{2mx}}{n.\cos{2nx}}$$

$$\cfrac{n.\sin{2mx}}{m.\sin{2nx}}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 2

$xy + x^2 y^2 = c$ ; then the value of

$\displaystyle \frac{dy}{dx}$ will be

0%
$\displaystyle \frac{x}{y}$
0%
$\displaystyle \frac{-y}{x}$
0%
$\displaystyle \frac{y}{x}$
0%
$\displaystyle \frac{-x}{y}$

Explanation

Given :

$xy + x^2 y^2 = c$ , differentiating with respect to x,

$\displaystyle \frac{d}{dx} (xy + x^2 y^2) = \frac{d}{dx} (c)$

or,

$\displaystyle \frac{d}{dx} (xy) + \frac{d}{dx} (x^2 y^2) = \frac{d}{dx} (c)$

or,

$\displaystyle \left ( x \frac{dy}{dx} +y \right ) + \left ( x^2.2y \frac{dy}{dx} + y^2 .2x \right ) = 0$

or,

$(x + 2x^2 y) \displaystyle \frac{dy}{dx} = - y - 2y^2 x$

$\therefore \displaystyle \frac{dy}{dx} = \frac{-y (1 + 2xy)}{x (1 + 2xy)} = \frac{-y}{x}.$

${x}^{2}+{y}^{2}=4$ , then

$y\cfrac { dy }{ dx } +x$ is equal to

0%
$4$
0%
$0$
0%
$1$
0%
$-1$

Explanation

Given

${x}^{2}+{y}^{2}=4$
On differentiating w.r.t

$x$ ,we get

$2x+2y\cfrac { dy }{ dx } =0$

$\Rightarrow$

$y\cfrac { dy }{ dx } +x=0$

$y$ is a function of

$x$ and

$\log { \left( x+y \right) } -2xy=0$ then the value of

$y'(0)$ is equal to

0%
$1$
0%
$-1$
0%
$2$
0%
$0$

Explanation

When

$x=0$ , we have

$\log { \left( 0+y \right) } -2\times 0\times y=0$

$\Rightarrow \quad \log { y } =0$

$\Rightarrow \quad y={ e }^{ 0 }=1$

Given equation is

$\log { (x+y) } -2xy=0$

On differentiating w,r,t

$x$ we get

$\cfrac { 1 }{ x+y } \left( 1+\cfrac { dy }{ dx } \right) -2y-2x\cfrac { dy }{ dx } =0$

On putting

$x=0, y=1$ we get

$\left( 1+\cfrac { dy }{ dx } \right) -2-0=0$

$\Rightarrow \left( 1+\cfrac { dy }{ dx } \right) -2=0$

$\Rightarrow \cfrac { dy }{ dx } =1$

$ie,\quad y'(x)=1\quad$

$\Rightarrow y'(0)=1$

Find the derivative of

$e^x+e^y=e^{x+y}$

0%
$-e^{x-y}$
0%
$e^{x-y}$
0%
$-e^{y-x}$
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$e^{y-x}$

Explanation

Given,

$e^x+e^y=e^{x+y}=e^xe^y$

$\Rightarrow e^{-y}+e^{-x}=1$
On differentiating, we get

$-e^{-y}\displaystyle\frac{dy}{dx}+e^{-x}(-1)=0$

$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{e^{-x}}{-e^{-y}}$

$\Rightarrow \displaystyle\frac{dy}{dx}=-e^{y-x}$

$y$ is expressed in terms of a variable

$x$ as

$y = f(x)$ , then

$y$ is called

0%
Explicit function
0%
Implicit function
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Linear function
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Identity function

Explanation

$y$ is expressed in terms of a variable

$x$ as

$y=f(x)$ , then

$y$ is called explicit function.

$y=\sqrt { \sin { x } +\sqrt { \sin { x } +\sqrt { \sin { x } +\dots \infty } } }$ , then

$\left( 2y-1 \right) \dfrac { dy }{ dx }$ is equal to

0%
$\sin { x }$
0%
$-\cos { x }$
0%
$\cos { x }$
0%
$-\sin { x }$

Explanation

Given,

$y=\sqrt { \sin { x } +\sqrt { \sin { x } +\sqrt { \sin { x } +\dots \infty } } }$

$\therefore y=\sqrt { \sin { x } +y }$

$\Rightarrow { y }^{ 2 }=\sin { x } +y$
On differentiating with respect to

$x$ , we get

$2y\dfrac { dy }{ dx } =\cos { x } +\dfrac { dy }{ dx }$

$\Rightarrow \left( 2y-1 \right) \dfrac { dy }{ dx } =\cos { x }$

The value of

$\dfrac{d}{dx}\{x(x-1)(x-2)(x-3)\}$ at

$x=3$

0%
$0$
0%
$3$
0%
$6$
0%
$24$

Explanation

Now,

$\dfrac{d}{dx}\{x(x-1)(x-2)(x-3)\}$

$=(x-3)\dfrac{d}{dx}\{x(x-1)(x-2)\}+$

$x(x-1)(x-2)\dfrac{d}{dx}\{(x-3)\}$ [ Using product rule of differentiation ]

$=(x-3)\dfrac{d}{dx}\{x(x-1)(x-2)\}+$

$x(x-1)(x-2).1$

$x=3$ the value of

$\dfrac{d}{dx}\{x(x-1)(x-2)(x-3)\}$ is

$0$

$=0+3.2.1.1$

$=6$ .

Differentiate

$2x^{3/2} + 2x^{5/2} +C$

0%
$\cfrac { dy }{ dx } =\sqrt { x } \left( 3+5x \right)$
0%
$\cfrac { dy }{ dx } =\sqrt { x } \left( 3-5x \right)$
0%
$\cfrac { dy }{ dx } =-\sqrt { x } \left( 3+5x \right)$
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None of these

Explanation

Let

$y=2{ x }^{ 3/2 }+2{ x }^{ 5/2 }+C$

As we know,

$\cfrac { d }{ dx } \left( { x }^{ n } \right) =n{ x }^{ n-1 }$

$\therefore \cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( 2{ x }^{ 3/2 } \right) +\cfrac { d }{ dx } \left( 2{ x }^{ 5/2 } \right)$

$\quad \quad =2.\cfrac { 3 }{ 2 } { x }^{ 1/2 }+2.\cfrac { 5 }{ 2 } { x }^{ 3/2 }$

$\therefore \cfrac { dy }{ dx } =\sqrt { x } \left( 3+5x \right)$

Find

$\dfrac{dy}{dx}$ of the given function

$y^{x} =x^{y}.$

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$y^x logy$ = $yx^{(y-1)}$
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$0$
0%
$x^ylogx$
0%
$x^y logy$ = $yx^{(y-1)}$

Explanation

Differentiating with respect to

$x$

$y^x\log y=yx^{(y-1)}$

The derivative of

$y = \left( {1 - x} \right)\left( {2 - x} \right)...\left( {n - x} \right)\,at\,x = 1$ is equal to :-

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$1$
0%
$\left( { - 1} \right)\left( {n - 1} \right)!$
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$n! - 1$
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${\left( { - 1} \right)^{n - 1}}\left( {n - 1} \right)!$

Find

$\dfrac{{dy}}{{dx}}$ of the following

$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$

0%
$x+2x^2+6x^3+(n-1)(n-2)x^{n-3}$
0%
$2+6x+(n-1)x^{n-2}$
0%
$2+6x+(n-1)(n-2)x^{n-3}$
0%
None

Explanation

$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$

$\dfrac{{dy}}{{dx}} = 2 + 6x + \left( {n - 1} \right)\left( {n - 2} \right){x^{n - 3}}$

The value of

$\dfrac{d}{dx}\displaystyle\int\limits_{2}^{x^2}(t-1)dt$

0%
$(x^2-1)$
0%
$x(x^2-1)$
0%
$2x(x^2-1)$
0%
none of these

Explanation

Given,

$\dfrac{d}{dx}\displaystyle\int\limits_{2}^{x^2}(t-1)dt$ .

Now applying the rule of differentiation under integral sign we get,

$\dfrac{d}{dx}\displaystyle\int\limits_{2}^{x^2}(t-1)dt=(x^2-1).2x-(2-1).0=2x(x^2-1).$

Let

$f(x)=x^2e^x$ then

$f''(0)=$

0%
$1$
0%
$0$
0%
$2$
0%
$7$

Explanation

Given,

$f(x)=x^2e^x$

Now differentiating both sides with respect to

$x$ we get,

$f'(x)=x^2e^x+2xe^x$

Again differentiating both sides with respect to

$x$ we get,

$f''(x)=x^2e^x+4xe^x+2e^x$ .

Now

$f''(0)=2$ .

The equation of the tangent to the curve

$y=x+\cfrac{4}{{x}^{2}}$ , that is parallel to the x-axis is

0%
$y=0$
0%
$y=1$
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$y=2$
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$y=3$

Explanation

Given tangent is parallel to x axis

so slope is 0

$\dfrac{dy}{dx}=0$

$1-\dfrac{8}{x^3}=0$

$x=2$

therefore at

$x=2$

$y=3$

equation of tangent is

$y=3$

$f(x)=\displaystyle \frac{x}{\sqrt{1-x^{2}}},g(x)=\frac{x}{\sqrt{1+x^{2}}}$ , then

$\displaystyle \frac{d}{dx}(fog (x))=$

0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

$\displaystyle f(g(x)) = \frac{g(x)}{\sqrt{1-(g(x))^2}}$ provided

$g(x) \neq 1$

$\displaystyle \quad = \frac{\dfrac{x}{\sqrt{1+x^2}}}{\sqrt{1-\dfrac{x^2}{1+x^2}}} =x$

$\therefore \dfrac{d}{dx}(fog(x)) = \dfrac{d}{dx}(x) =1$

Let

$y=\sqrt{(\sin x+\sin 2x+\sin 3x)^2+(\cos x+\cos 2x+\cos 3x)^2}$ then which of the following(s) is correct?

0%
$\dfrac{dy}{dx}$ when $x=\dfrac{\pi}{2}$ is $-2$
0%
Value of y when $x=\dfrac{\pi}{5}$ is $\dfrac{3+\sqrt{5}}{2}$
0%
Value of y when $x=\dfrac{\pi}{12}$ is $\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{2}$
0%
y simplifies to $(1+2\cos x)$ in $[0, \pi]$

Explanation

$y=\sqrt{(\sin x+\sin 2 x+\sin 3 x)^2+(\cos x+\cos 2 x+\cos 3 x)^2}$

$=\sqrt{(\sin ^2 x+\cos^2 x)+(\sin ^2 2 x+\cos^2 2 x)+(\sin ^2 3 x+\cos^2 3 x)+2(\sin 2 x\sin x+\cos 2 x\cos x)+2(\sin 2 x \sin 3 x+\cos 2 x\cos 3 x)+2(\sin x\sin 3 x+\cos x\cos 3 x}$

$=\sqrt{1+1+1+2\cos(2 x-x)+2\cos (3 x-2 x)+2\cos (3 x- x)}=\sqrt{3+4 \cos x+2\cos 2 x}=\sqrt{4\cos^2 x+4\cos x+1}$

$=|2\cos x+1|$

$x=\dfrac{\pi}{2},\dfrac{d y}{d x}=-2\sin x=-2\sin \pi/2=-2$

$x=\dfrac{\pi}{5},y=2\cos \dfrac{\pi}{5}+1=2\times\dfrac{\sqrt{5}+1}{4}+1=\dfrac{\sqrt{5}+3}{2}$

$x=\dfrac{\pi}{12},y=2\cos \dfrac{\pi}{12}+1=2\times \dfrac{\sqrt{3}+1}{2\sqrt{2}}+1=\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{\sqrt{2}}$

$\dfrac{d}{{dx}}\left( {\dfrac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}} \right) = ax + b,\,then\left( {a,b} \right) =$

0%
$(-1,2)$
0%
$(-2,1)$
0%
$(2,-1)$
0%
$(1,2)$

Explanation

Given expression,

$1+x^2+x^4=(x^2+1-x)(x^2+1+x)$

$=>\dfrac{1+x^2+x^4}{1+x+x^2}=1-x+x^2$

$=>\dfrac{d(1-x+x^2)}{dx}=2x-1$

Comparing with coefficient,

$a=2,b=-1$

$\sin ^{ 2 }{ mx } +\cos ^{ 2 }{ ny } ={ a }^{ 2 }$ then

$\cfrac{dy}{dx}=$

0%
$\cfrac{m.\sin{2mx}}{n.\sin{2ny}}$
0%
$\cfrac{m.\sin{mx}}{n.\sin{nx}}$
0%
$\cfrac{-m.\cos{2mx}}{n.\cos{2nx}}$
0%
$\cfrac{n.\sin{2mx}}{m.\sin{2nx}}$

Explanation

differentiate on both sides

$2m\sin{mx}\cos{mx}(dx)-2n\cos{ny}\sin{ny}(dy)=0$

$(\dfrac{dy}{dx})=\dfrac{m.\sin{2mx}}{n.\sin{2ny}}$

Find

$\dfrac{dy}{dx}$ of function

$y= e^{x^3} +\dfrac{1}{2} \log x$

0%
$2.e^{x^3}x^2+\dfrac {1}{2x}$
0%
$e^{x^3}x^2+\dfrac {1}{2x}$
0%
$3.e^{x^3}x^2+\dfrac {1}{2x}$
0%
$3.e^{x^3}x^2+\dfrac {1}{x}$

Explanation

Given y= $e^{x^3} +\frac{logx}{x}$

$(\frac{dy}{dx})=(\frac{de^{x^3}}{dx})+(\frac{d(\frac{logx}{2})}{dx})\\=(\frac{de^{x^3}}{dx^3})\times(\frac{dx^3}{dx})+(\frac{1}{2})(\frac{dlogx}{dx})\\=3x^2e^{x^3}+(\frac{1}{2x})$

$x^{\frac{1}{2}} + 1= t$
differentiate w.r.t. x

0%
$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{4\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{8\sqrt{x}}$
0%
$\dfrac{dt}{dx} = \dfrac{1}{16\sqrt{x}}$

Explanation

$x^{\frac{1}{2}} + 1= t$

$\dfrac{dt}{dx} = \dfrac{1}{2}(x) ^{\frac{1}{2}-1} + 0$

$\frac{1}{2}(x) ^{\frac{-1}{2}}$

$\dfrac{1}{2x^{\frac{1}{2}}}$

$\dfrac{dt}{dx} = \dfrac{1}{2\sqrt{x}}$

Differentiate:

$x^{100} + \sin x - 1$

0%
$100x^{99}-\cos x$
0%
$100x^{99}+\cos x$
0%
$x^{99}+\cos x$
0%
$100x^{99}+\sin x$

Explanation

$\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} +\sin x -1 \right )$

$=\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( x^{100} \right )+\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( \sin x \right )-\cfrac{\mathrm{d} }{\mathrm{d} x}\left ( 1 \right )$

$=100x^{100-1}+\cos x -0$

$=100x^{99}+\cos x$

$f(x)=x\log x$ , then find f'(x).

0%
$1+x\log x$
0%
$1+x$
0%
$x+\log x$
0%
$1+\log x$

Explanation

We have,

$f(x)=x\log x$

On differentiating w.r.t

$x$ , we get

$f'(x)=\log x\times 1+x\times \dfrac{1}{x}$

$f'(x)=\log x+1$

Hence, this is the answer.

$\displaystyle\int \dfrac{f(x)dx}{\log(\sin x)}$

$=\log (\log(\sin x))$ then

$f(x) =$

0%
$sinx$
0%
$cosx$
0%
$\log(sinx)$
0%
$cotx$

Explanation

$\int \frac{f(x)}{log (sin x)}dx= log [log(sin x)]$

on differentiating both side cort x we get

$\frac{d}{dx}\int \frac{f(x)}{log(sin x)} dx=\frac{d}{dx}(log (log sin x))$

$\frac{f(x)}{log (sin x)}=\frac{cos x}{log (in x)}$

$[\therefore \frac{d}{dx}log k=\frac{1}{k}]$

$[f(x)= cos x]$

1199975_1382341_ans_7b260c5913cf4deba5e709f74561da6c.JPG

$y=5x^2+8x$ find

$\dfrac {dy}{dx}$

0%
$10x+8$
0%
$5x+8$
0%
$10x^2+8x$
0%
none of these

Explanation

The Equation is

$y=5x^2+8x$

$\dfrac {dy}{dx}=\dfrac d{dx}(5x^2+8x)\\\dfrac{dy}{dx}=10x+8$

$y=\log \sin x$ find

$x$ if

$y=0$

0%
$\dfrac {\pi}2$
0%
$\pi$
0%
${\dfrac {\pi}3}$
0%
$\dfrac {-\pi}2$

Explanation

$y=\log \sin x\\y=0\\\log \sin x=\log 1\\\sin x=1\\\sin x=\sin \dfrac{\pi}2\\x=\dfrac{\pi}2$

$\mathrm{f}(\mathrm{x})=0$ has a repeated root

$\alpha$ , then another equation having

$\alpha$ as root, is

0%
$\mathrm{f}(2\mathrm{x})=0$
0%
$\mathrm{f}(3\mathrm{x})=0$
0%
$\mathrm{f}^{'}(\mathrm{x})=0$
0%
$\mathrm{f}^{''}(\mathrm{x})=0$

Explanation

Repeated roots mean that the curve will touch the

$x$ -axis.
The curve will have a convex or concave part at the repeated root

$\alpha$ or a point of inflection.
In any case

$f'(x)$ will always be zero at the point of repeated root

$\alpha$ .

$[\mathrm{x}]$ denotes the greatest integer contained in

$\mathrm{x},$ then for 4

$<\mathrm{x}<5,\ \displaystyle \frac{d}{dx}\{[x]\}=$

0%
$[x - 4, 5]$
0%
$[x]$
0%
$0$
0%
$1$

Explanation

$[\ ]$ denote greatest integer function ,i.e

$[n+f]=n$ ,

where

$n=integer\$ and

$f= fraction$ ,

$4<x<5$ , then

$[x] =4 =$ constant
Hence

$\dfrac{d}{dx}\{[x]\} =\dfrac{d}{dx}(4)=0$

$y=\log{(x+\sqrt{1+x^2})}$ , then

$y^{''}(0)$ is

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

$y=\log(x+\sqrt{1+x^{2}})$

${y}'=\dfrac{1+\dfrac{x}{\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}}=\dfrac{1}{\sqrt{1+x^{2}}}$

${y}''=\dfrac{-1}{2}(\dfrac{2x}{(1+x^{2})^{3/2}})$

${y}''(0)=0$

$\displaystyle \mathrm{A}:\dfrac{d}{dx}(\sin x)\ at\ x=\frac{\pi}{2}$

$\displaystyle \mathrm{B}:\dfrac{d}{dx}(\tan^{-1}{x})$ at

${x}=1$

$\displaystyle \mathrm{C}:\dfrac{d}{dx}(\mathrm{e}^{x})$ at

${x}=0$

$\mathrm{D}:\dfrac{d}{dx}(x^{x})\ at\ x=e$

Arrangement of the above values in the increasing order of the magnitude

0%
B, C, A, D
0%
D, A, B, C
0%
D, B, C, A
0%
A, B, C, D

Explanation

$A.\dfrac{d}{dx}(sinx) = cosx$

Put

$x=\dfrac{\pi }{2}$

$=0$

$B.\dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^{2}}$

Put

$\ x=1$

$=\dfrac{1}{2}$

$C. \dfrac{d}{dx}(e^{x})=e^{x}$

Put

$x=0$

$=1$

$D.\dfrac{d}{dx}(x^{x})=\dfrac{d}{dx}(e^{xLogx})=x^{x}(Log x+1)$

$=2e^{e}$

$y=\displaystyle \log{\left(\frac{1+x}{1-x}\right)^{\frac{1}{4}}}-\frac{1}{2}\tan^{-1}(x)$ , then

$\displaystyle \frac{dy}{dx}=$

0%
$\displaystyle \frac{x}{1-x^{2}}$
0%
$\displaystyle \frac{x^{2}}{1-x^{4}}$
0%
$\displaystyle \frac{x}{1+x^{4}}$
0%
$\displaystyle \frac{x}{1-x^{4}}$

Explanation

$y=\dfrac{1}{4}(\log(1+x)-\log(1-x))-\dfrac{1}{2}\tan^{-1}x$

$\dfrac{dy}{dx}=\dfrac{1}{4}\left(\dfrac{2}{(1-x^2)}\right)-\dfrac{1}{2(1+x^{2})}$
=

$\dfrac{1}{4}\left(\dfrac{2}{1-x^{2}}\right)-\dfrac{1}{2}\dfrac{1}{(1+x^{2})}$
=

$\dfrac{1}{2}\left(\dfrac{1+x^{2}-(1-x^{2})}{1-x^{4}}\right)$
=

$\dfrac{x^{2}}{1-x^{4}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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