If $$x$$ $$sin$$ $$y$$$$=3 sin y$$ $$+$$ $$4 cos y$$, then $$\frac { dy }{ dx } =$$

$$\frac { { -cos }^{ 2 }y }{ 4 } $$

$$\frac { { cos }^{ 2 }y }{ 4 } $$

If $$x\sqrt {1+y}+y\sqrt {1+x}=0$$, then $$\dfrac {dy}{dx}=$$

$$-\dfrac {\sqrt {y+1}}{\sqrt {x+1}}$$

$$\dfrac {\sqrt {1+y}}{\sqrt {1+x}}$$

If $$(\cos x)^{y}=(\sin y)^{x}$$, then $$\dfrac{dy}{dx}=$$

$$\dfrac{\log (\sin y)-y \tan x}{\log (\cos x)+ x \cot y}$$

$$\dfrac{\log (\sin y)+y \tan x}{\log (\cos x)- x \cot y}$$

$$\dfrac{\log (\sin y)}{\log (\cos x)}$$

$$\dfrac{\log (\cos x)}{\log (\sin y)}$$

If $$y=(x+\sqrt{x^{2}+a^{2}})^{n}$$ then $$\dfrac{dy}{dx}=$$

$$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$$

$$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$$

If $$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$$ then $$\frac {dy}{dx}=$$

$$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$$

$${e^{{{\tan }^2}x}}{\sec ^2}x$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 11

$x+y=\sin(x+y)$ then

$\dfrac{dy}{dx}$ =

0%
$\dfrac{1}{2}$
0%
$0$
0%
$-1$
0%
$\dfrac{1}{3}$

$\sqrt { 1 - x ^ { 4 } } + \sqrt { 1 - y ^ { 4 } } = a \left( x ^ { 2 } - y ^ { 2 } \right)$ , then

$\frac { d y } { d x } = \frac { x } { y } \sqrt { \frac { 1 - y ^ { 4 } } { 1 - x ^ { 4 } } }$ .

0%
True
0%
False

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

$\dfrac{dy}{dx}$ at the point

$(0,2)$ is:

0%
$0$
0%
$-32$
0%
$4$
0%
$-2$

Explanation

$x^2+y^2=4$

Differentiating both sides w.r.t.

$x,$ we get:

$\Rightarrow$

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

$\Rightarrow$

$2y\dfrac{dy}{dx}=-2x$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{-2x}{2y}$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{-x}{y}$

$\Rightarrow$

$\left(\dfrac{dy}{dx}\right)_{(0,2)}=\dfrac{0}{2}$

$\therefore$

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

$x$

$sin$

$y$

$=3 sin y$

$+$

$4 cos y$ , then

$\frac { dy }{ dx } =$

0%
$\frac { { -sin }^{ 2 }y }{ 4 }$
0%
$\frac { { sin }^{ 2 }y }{ 4 }$
0%
$\frac { { -cos }^{ 2 }y }{ 4 }$
0%
$\frac { { cos }^{ 2 }y }{ 4 }$

Explanation

$x\sin{y}=3\sin{y}+4\cos{y}$

$\Rightarrow\,\left(x-3\right)\sin{y}=4\cos{y}$

$\Rightarrow\,x-3=4\dfrac{\cos{y}}{\sin{y}}$

$\Rightarrow\,x-3=4\cot{y}$

Differentiating both sides w.r.t

$x$

$\Rightarrow\,1=4\left(-{\csc}^{2}{y}\right)\dfrac{dy}{dx}$

$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{-1}{4{\csc}^{2}{y}}$

$\therefore\,\dfrac{dy}{dx}=\dfrac{-{\sin}^{2}{y}}{4}$

$x+y=t-\dfrac{1}{t},x^{2}+y^{2}=t^{2}+\dfrac{1}{t^{2}},$ then

$\dfrac{dy}{dx}$ is equal to

0%
$\dfrac{-y}{x}$
0%
$-\dfrac{1}{x}$
0%
$\dfrac{1}{x^{2}}$
0%
$-\dfrac{1}{x^{2}}$

Explanation

We have,

$x+y=t-\dfrac{1}{t}\,\,......\,\,\left( 1 \right)$

${{x}^{2}}+{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}\,\,.......\,\,\left( 2 \right)$

By equation (2) to,

${{x}^{2}}+{{y}^{2}}={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2+2$

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

${{x}^{2}}+{{y}^{2}}-2={{\left( t-\dfrac{1}{t} \right)}^{2}}\,\,.......\,\,\left( 3 \right)$

By equation (1) and (3) to, and we get,

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

${{x}^{2}}+{{y}^{2}}-2={{x}^{2}}+{{y}^{2}}+2xy$

$-2=2xy$

$xy=-1$

On differentiating and we get,

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

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

$x\dfrac{dy}{dx}=-y$

$\dfrac{dy}{dx}=\dfrac{-y}{x}$

Hence, this is the answer.

$X={ { { { e }^{ y+e } }^{ y+e } }^{ y+e } }^{ y+...\infty }$ , then

$\dfrac {dy}{dx}$ is

0%
$\dfrac {X}{1+X}$
0%
$\dfrac {1}{X}$
0%
$\dfrac {1-X}{X}$
0%
None of these

$a{x^2} + 2hxy + b{y^2} = 0$ then

$\frac{{dy}}{{dx}}$ is equal to

0%
$\frac{y}{x}$
0%
$\frac{x}{y}$
0%
$- \frac{x}{y}$
0%
none of these

$y=a\ \sin\ x+b\ \cos\ x$ , then

$y^{2}+\left ( \dfrac{dy}{dx} \right )^{2}$ is

0%
function of $x$
0%
function of $y$
0%
function of $x$ and $y$
0%
constant

Explanation

$y=a\sin{x}+b\cos{x}$

$\dfrac{dy}{dx}=a\cos{x}-b\sin{x}$

${y}^{2}+{\left(\dfrac{dy}{dx}\right)}^{2}={\left(a\sin{x}+b\cos{x}\right)}^{2}+{\left(a\cos{x}-b\sin{x}\right)}^{2}$

$={a}^{2}{\sin}^{2}{x}+{b}^{2}{\cos}^{2}x+2ab\sin{x}\cos{x}+{a}^{2}{\cos}^{2}{x}+{b}^{2}{\sin}^{2}x-2ab\sin{x}\cos{x}$

$={a}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{b}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)$

$={a}^{2}+{b}^{2}$ since

${\sin}^{2}{x}+{\cos}^{2}{x}=1$

$=constant$

$y=y(x)$ and it follows the relation

$e^{xy^{2}}+y\cos(x^{2})=5$ then

$y'(0)$ is equal to

0%
$4$
0%
$-16$
0%
$-4$
0%
$16$

$y=a^{{a}^{{x}}}$ , then

$\dfrac {dy}{dx}=$

0%
$y.a^{x}{(\log a)}^{2}$
0%
$y.a^{x}.\log a$
0%
${(y.a^{x})}^{2}$
0%
$(y.a^{x})$

Explanation

$\quad y={ a }^{ { x }^{ y } }$

taking logarithm on both sides

$\log { y } ={ x }^{ y }\log { a }$

again we take logarithm on both sides

$\log { \log { y } } =y\left( \log { x } \right) +\log { \log { a } }$

taking derivative on both sides w.r. t

$x$

$\cfrac { 1 }{ \log { y } } .\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { dy }{ dx } \log { x } +y.\cfrac { 1 }{ x } +0$

$\Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1 }{ y\log { y } } -\log { x } \right) =\cfrac { y }{ x } \Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1-y\log { x } .\log { y } }{ y\log { y } } \right) =\cfrac { y }{ x }$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { { y }^{ 2 }\log { y } }{ x\left( 1-y\log { x } .\log { y } \right) }$

Let

$f$ be a differentiable function satisfying the condition

$f(\dfrac{x}{y})=\dfrac{f(x)}{f(y)}$ , for all

$x,y\neq 0\ \epsilon \ R$ and

$f(y)\neq 0$ . If

$f'(1)=2$ then

$f'(x)$ is equal to

0%
$2f(x)$
0%
$\dfrac{2f(x)}{x}$
0%
$2xf(x)$
0%
$\dfrac{2f(x)}{2}$

$f\left( x \right) =\sqrt { 1+\sqrt { 6+\sqrt { 5+{ x }^{ 2 } } } }$ , then

$f'(2)=$

0%
$1$
0%
$\dfrac {1}{12}$
0%
$\dfrac {1}{36}$
0%
$None$

Let

$f(x)$ be differentiable function such that

$f\left(\dfrac{x+y}{1-xy}\right)=f(x)+f(y) \forall x$ and

$y$ . If

$\underset { x\rightarrow 0 }{ lt } \dfrac { f\left( x \right) }{ x } =\dfrac { 1 }{ 3 }$ then

$f(1)$ equals

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{12}$
0%
$\dfrac{1}{8}$

Explanation

We have,

$f\left( {\dfrac{{x + y}}{{1 - xy}}} \right) = f\left( x \right) + f\left( y \right)\,\,\,\forall {x^2}y$

$\Rightarrow f\left( x \right) = k{\tan ^{ - 1}}\left( x \right)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{k{{\tan }^{ - 1}}x}}{x} = \dfrac{1}{3}$

$k = \dfrac{1}{3}$

$f\left( x \right) = \dfrac{1}{3}{\tan ^{ - 1}}x$

$f\left( 1 \right) = \dfrac{1 }{{12}}$

Hence, this is the answer.

$f(x+y)=2f(x).f(y)$ for all

$x,y$ , where

$f'(0)=3$ and

$f'(4)=2$ then

$f'(4)=3$ is equal to

0%
6
0%
12
0%
4
0%
3

$y = {\log _{10}}\left( {\sin x} \right),$ then

$\dfrac{dy}{dx}$ equals to:

0%
$\sin x\,{\log _{10}}e$
0%
$\cos x\,{\log _{10}}e$
0%
$\cot x\,{\log _{10}}e$
0%
$\cot x$

Explanation

$y=log_{10}\sin (x)$

$\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x$

$\dfrac{dy}{dx}=\cot x..........$

$\because log 10=1$ .

$x\sqrt {1+y}+y\sqrt {1+x}=0$ , then

$\dfrac {dy}{dx}=$

0%
$-\dfrac {1}{(x+1)^{2}}$
0%
$\dfrac {1}{(x+1)^{2}}$
0%
$-\dfrac {\sqrt {y+1}}{\sqrt {x+1}}$
0%
$\dfrac {\sqrt {1+y}}{\sqrt {1+x}}$

Explanation

$x\sqrt{1+y}+y\sqrt{1+x}=0$

$=x\sqrt{1+y}=-y\sqrt{1+x}$

Squaring,

$x^2(1+y)=y^2(1+x)$

$=x^2+x^2y=y^2+xy^2$

$=x^2-y^2=xy^2-x^2y$

$=x^2-y^2=xy(y-x)$

$=(x+y)(x-y)=xy(y-x)$

$=(x+y)=-xy$

$=x+xy+y=0$

$=(1+x)y=-x$

$y=-x/(1+x)$

differentiating,

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

$y = \sin x \left[ \frac { 1 } { \sin x \cdot \sin 2 x } + \frac { 1 } { \sin 2 x \sin 3 x } + \ldots\right.$

$+ \frac { 1 } { \sin n x \sin ( n + 1 ) x } ]$ then

$\frac { d y } { d x } =$

0%
$\cot x - \cot ( n + 1 ) x$
0%
$( n + 1 ) \cos e c ^ { 2 } ( n + 1 ) x - \cos e c ^ { 2 } x$
0%
$\csc ^ { 2 } x - ( n + 1 ) \cos e c ^ { 2 } ( n + 1 ) x$
0%
$\cot x + \cot ( n + 1 ) x$

Explanation

$\begin{array}{l} y=\left[ { \frac { 1 }{ { \sin x\sin 2x } } +\frac { 1 }{ { \sin 2x\sin 3x } } +...........+\frac { 1 }{ { \sin nx\sin \left( { n+1 } \right) x } } } \right] \\ y=\left[ { \frac { { \sin x } }{ { \sin x\sin 2x } } +\frac { { \sin x } }{ { \sin 2x\sin 3x } } +...........+\frac { { \sin x } }{ { \sin nx\sin \left( { n+1 } \right) x } } } \right] \\ y=\left[ { \frac { { \sin \left( { 2x-x } \right) } }{ { \sin 2x\sin x } } +\frac { { \sin \left( { 3x-2x } \right) } }{ { \sin 3x\sin 2x } } +...........+\frac { { \sin \left( { n+1 } \right) x-nx } }{ { \sin \left( { n+1 } \right) x\sin nx } } } \right] \\ y=\left[ { \frac { { \sin \left( { 2x } \right) \cos x-\cos 2x\sin x } }{ { \sin 2x\sin x } } +\frac { { \sin \left( { 3x } \right) \cos 2x-\cos 3x\sin 2x } }{ { \sin 3x\sin 2x } } +...........+\frac { { \sin \left( { n+1 } \right) x\cos nx-\cos \left( { n+1 } \right) x\sin x } }{ { \sin \left( { n+1 } \right) x\sin nx } } } \right] \\ y=\left[ { \frac { { \sin \left( { 2x } \right) \cos x } }{ { \sin 2x\sin x } } -\frac { { \cos 2x\sin x } }{ { \sin 2x\sin x } } +\frac { { \sin \left( { 3x } \right) \cos 2x } }{ { \sin 2x\sin x } } \frac { { -\cos 3x\sin 2x } }{ { \sin 3x\sin 2x } } +...........+\frac { { \sin \left( { n+1 } \right) x\cos nx } }{ { \sin \left( { n+1 } \right) x\sin nx } } -\frac { { \cos \left( { n+1 } \right) x\sin x } }{ { \sin \left( { n+1 } \right) x\sin nx } } } \right] \\ y=\left[ { \frac { { \cos x } }{ { \sin x } } -\frac { { cos2x } }{ { \sin 2x } } +\frac { { cos2x } }{ { \sin 2x } } -\frac { { cos3x } }{ { \sin 3x } } +.....+\frac { { \cos nx } }{ { \sin nx } } -\frac { { \cos \left( { n+1 } \right) x } }{ { \sin \left( { n+1 } \right) x } } } \right] \\ y=\left[ { \cot x-\cot 2x+\cot 2x-\cot 3x+...\cot nx-\cot \left( { n+1 } \right) x } \right] \\ y=\left[ { \cot x-\cot \left( { n+1 } \right) x } \right] \\ \frac { { dy } }{ { dx } } =-\cos e{ c^{ 2 } }x-\left( { -\cos e{ c^{ 2 } }\left( { n+1 } \right) x } \right) \left( { n+1 } \right) \\ \frac { { dy } }{ { dx } } =\left( { n+1 } \right) \cos e{ c^{ 2 } }\left( { n+1 } \right) x-\cos e{ c^{ 2 } }x \end{array}$

$(\cos x)^{y}=(\sin y)^{x}$ , then

$\dfrac{dy}{dx}=$

0%
$\dfrac{\log (\sin y)+y \tan x}{\log (\cos x)- x \cot y}$
0%
$\dfrac{\log (\sin y)-y \tan x}{\log (\cos x)+ x \cot y}$
0%
$\dfrac{\log (\sin y)}{\log (\cos x)}$
0%
$\dfrac{\log (\cos x)}{\log (\sin y)}$

Explanation

Given,

$(\cos x)^y=(\sin y)^x$

Now taking logarithm with respect to the base

$e$ we get,

$y\log (\cos x)=x\log(\sin y)$

Now differentiating both sides with respect to

$x$ we get,

$\log(\cos x)\dfrac{dy}{dx}+y.\dfrac{(-\sin x)}{\cos x}$

$=\log(\sin y)+x.\dfrac{(\cos y)}{\sin y}\dfrac{dy}{dx}$

or,

$(\log(\cos x)-x\cot y)\dfrac{dy}{dx}=\log (\sin y)+y\tan x$

or,

$\dfrac{dy}{dx}=\dfrac{\log (\sin y)+y\tan x}{\log(\cos x)-y\cot x}$

Let y be an implicit function of

$x$ defined by

$x^{2x}-2x^x\cot\:y-1=0$ . Then

$y'\left(1\right)$ equals

0%
$-1$
0%
$1$
0%
$\log\:2$
0%
$-\log\:2$

Explanation

Now

$x=1\Rightarrow 1-2\cot{y}-1=0$

$\Rightarrow \cot{y}=0$ or

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

Let

$u,v,w$ be the functions of

$x$ then

$\Rightarrow u+v+w=0$ and

$\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}=0$

Take

$u={x}^{2x}$

$\Rightarrow \log{u}=\log{{x}^{2x}}=2x\log{x}$ using the power form of logarithm

$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=2\left[x\times\dfrac{1}{x}+\log{x}\right]=2\left(1+\log{x}\right)$

$\therefore \dfrac{du}{dx}=2u\left(1+\log{x}\right)=2{x}^{2x}\left(1+\log{x}\right)$

Take

$v=-2{x}^{x}coty$

$\dfrac{dv}{dx}=-2\left[{x}^{2}\left(-{\csc}^{2}{y}\right)\dfrac{dy}{dx}+\cot{y}\dfrac{d{\left({x}^{x}\right)}}{dx}\right]$

$=-2\left[{x}^{2}\left(-{\csc}^{2}{y}\right)\dfrac{dy}{dx}+{x}^{x}\cot{y}\left(1+\log{x}\right)\right]$

Take

$w=-1$

$\dfrac{dw}{dx}=\dfrac{d\left(-1\right)}{dx}=0$ since differential of a constant

$=0$

Now,

$\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}=0$

$\Rightarrow 2{x}^{2x}\left(1+\log{x}\right)-2\left[{x}^{2}\left(-{\csc}^{2}{y}\right)\dfrac{dy}{dx}+{x}^{x}\cot{y}\left(1+\log{x}\right)\right]+0=0$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{2{x}^{2x}\left(1+\log{x}\right)-2{x}^{2}\left(1+\log{x}\right)\cot{y}}{-2{x}^{2}{\csc}^{2}{y}}$

$\Rightarrow \left[\dfrac{dy}{dx}\right]_{\left(1,\frac{\pi}{2}\right)}=\dfrac{2 \times {1}^{2}\left(1+\log{1}\right)-2\times {1}^{2}\left(1+\log{1}\right)\cot{\frac{\pi}{2}}}{-2\times {1}^{2}{\csc}^{2}{\frac{\pi}{2}}}$

$\Rightarrow {y}^{\prime}{\left(1\right)}=\dfrac{1\left(1+\log{1}\right)}{-2}{\sin}^{2}{\frac{\pi}{2}}$ since

$\dfrac{1}{{\csc}^{2}{\frac{\pi}{2}}}={\sin}^{2}{\frac{\pi}{2}}$

$\therefore {y}^{\prime}{\left(1\right)}=-1\times 1=-1$ since

$\sin{\dfrac{\pi}{2}}=1$

$y=(x+\sqrt{x^{2}+a^{2}})^{n}$ then

$\dfrac{dy}{dx}=$

0%
$y$
0%
$ny$
0%
$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$
0%
$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$

Explanation

$y={ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n }\quad$

$\cfrac { dy }{ dx } =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\cfrac { d }{ dx } \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right)$

$=n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\left( 1+\cfrac { x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right) =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }\left( \cfrac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } +x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right)$

$=\cfrac { n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n } }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\cfrac { n.y }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } }$

$x^m\cdot y^n=\left(x+y\right)^{m+n}$ , then

$\dfrac{dy}{dx}$ is ?

0%
$\dfrac{y}{x}$
0%
$\dfrac{x+y}{xy}$
0%
xy
0%
$\dfrac{x}{y}$

Explanation

We have

${x}^{m}{y}^{n}={\left(x+y\right)}^{m+n}$

$\log{{x}^{m}{y}^{n}}=\log{{\left(x+y\right)}^{m+n}}$

$\rightarrow$ by taking

$\log$ on both sides

$\log{{x}^{m}}+\log{{y}^{n}}=\log{{\left(x+y\right)}^{m+n}}$

$\because\log{ab}=\log{a}+\log{b}$

$m\log{x}+n\log{y}=\left(m+n\right)\log{\left(x+y\right)}$

$\because\log{{a}^{m}}=m\log{a}$

Differentiating both sides w.r.t

$x$ we get

$m\dfrac{1}{x}+n\dfrac{1}{y}\dfrac{dy}{dx}=\left(m+n\right)\dfrac{1}{\left(x+y\right)}\left[1+\dfrac{dy}{dx}\right]$

$\Rightarrow \dfrac{m}{x}-\dfrac{m+n}{x+y}=\left(\dfrac{m+n}{x+y}-\dfrac{n}{y}\right)\dfrac{dy}{dx}$ by seperating the terms

$\Rightarrow \dfrac{mx+my-mx-nx}{x\left(x+y\right)}=\left(\dfrac{my+ny-nx-ny}{y\left(x+y\right)}\right)\dfrac{dy}{dx}$

$\Rightarrow \dfrac{my-nx}{x}=\left(\dfrac{my-nx}{y}\right)\dfrac{dy}{dx}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}$

$2^{x}-2^{y}=2^{x+y}$ then

$\dfrac{dy}{dx}=$

0%
$2^{y-x}$
0%
$2^{y/x}$
0%
$-2^{y/x}$
0%
$2^{x/y}$

Explanation

since

$2^x - 2^y = 2^{x+y}$

hence

$2^x - 2^{x+y} = 2^y$

and

$2^y + 2^{x+y} = 2^x$

replace this in above we get

$\dfrac{dy}{dx} = \dfrac{2^y}{2^x}$

$\dfrac{dy}{dx} = 2^{y-x}$

hence this is the answer

1190137_1241190_ans_bb6605a7b4f845d6b50947f790c7b0d4.png

$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$ then

$\frac {dy}{dx}=$

0%
${e^{{{\tan }^2}x}}$
0%
${e^{{{\tan }^2}x}}{\sec ^2}x$
0%
$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$
0%
none

$y = {x^2} + \dfrac{1}{{{x^2} + \frac{1}{{{x^2} + \ldots \ldots \infty }}}},$ then

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

0%
$\dfrac{{ - x{y^2}}}{{{y^2} + 1}}$
0%
$\dfrac{{ 2xy}}{{2y-x^2}}$
0%
$\dfrac{{ - {x^2}{y^2}}}{{{y^2} + 1}}$
0%
$\dfrac{{ x{y^2}}}{{{y^2} + 1}}$

Explanation

$y = x^2 + \dfrac{1}{x^2 + \dfrac{1}{x^2 + ... \infty}}$

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

$y^2 = x^2 y + 1$

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

$\dfrac{dy}{dx} = \dfrac{2xy}{2y - x^2}$

$x\dfrac{dy}{dx}=y(\log y-\log x +1)$ , then the solution of the equation

0%
$\log\left(\dfrac{x}{y}\right)=cy$
0%
$\log\left(\dfrac{y}{x}\right)=cx$
0%
$x \log\left(\dfrac{x}{y}\right)=cy$
0%
$y \log\left(\dfrac{x}{y}\right)=cy$

$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$ , then

$\dfrac {dy}{dx}$ at

$x=1$ is equal to

0%
$\dfrac {1}{\sqrt 2}$
0%
$\dfrac {1}{2}$
0%
$1$
0%
$\sqrt 2$

If y(x) is the solution of the differential equation

$\left( x+2 \right) \dfrac { dy }{ dx } ={ x }^{ 2 }+4x-9,x\neq -2$ and y(0)=0, then y(-4) is equal to :

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

Explanation

$y(x)$ is the solution of the

$\triangle .E$

$(x+2)\dfrac{dy}{dx}=x^2+4x-9,x\neq-2$

$y(0)=0$

To find :

$y(-4)=?$

Solution : Solving the D.E

$=dy\dfrac{x^2+4x-9}{x+2}dx$

$dy=\dfrac{x^2+4x+4-13}{x+2}dx$

$dy=\dfrac{(x^2+4x+4)-13}{x+2}dx\rightarrow dy=\dfrac{(x+2)^2-13}{x+2}dx$

$dy=\left(x+2-\dfrac{13}{x+2}\right)dx$

Integrating both sides

$\displaystyle \int dy=\int \left[(x+2)-\dfrac{13}{x+2}\right]dx$

$y^x=\dfrac{x^2}{2}+2x-13 ln|x+2|+c$

Now

$y(0)=0$

$0=\dfrac{0}{2}+2(0)-13ln|0+2|+c$

$0=-13ln|2|+c$

$c=13 ln 2$

So,

$y(x)=\dfrac{x^2}{2}+2x-13 ln|x+2|+13 ln2$

$\dfrac{x^2}{2}+2c+13[ln 2-ln|x+2|]$

$=\dfrac{x^2}{2}+2x+13 ln\left|\dfrac{2}{x+2}\right|$

$y(-4)=\dfrac{(-4)^2}{2}+2(-4)+13ln\left|\dfrac{2}{4+2}\right|$

$=8-8+!3ln|-1|$

$=13\, ln 1=13(0)$

$=\boxed{0}$

The solution of the differential equation

$\left( \dfrac { d y } { d x } \right) ^ { 2 } - 3 x \left( \dfrac { d y } { d x } \right) - 2 y = 8$ is

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$y = 2 x ^ { 2 } + 4$
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$y = 2 x ^ { 2 } - 4$
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$y = 2 x + 4$
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$y = 2 x - 4$

Explanation

By option Verification

Put

$\dfrac{dy}{dx}=4x$

substitute this in the equation

$\implies y=2x^2-4$

A curve in the

$1^{st}$ quadrant passes through

$(1,1)$ . Its drifferential equation is

$(y-xy^{2})dx+(x+x^{2}y^{2})dy=0$ . Hence the equation of the curve is

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$y-\dfrac{1}{xy}=\ln y$
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$y-\dfrac{1}{xy}=\ln x$
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$y-xy=\ln y$
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$y-xy=\ln x$

$\dfrac {dy}{dx}=(e^ {y}-x)^ {-1}$ where

$y(0)=0$ then

$y$ is expressed explicity as

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$0.5\log_{e}(1+x^{2})$
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$\log_{e}(1+x^{2})$
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$\log _{ e } \left( x+\sqrt { 1+{ x }^{ 2 } } \right)$
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$\\ \log _{ e } \left( x+\sqrt { 1-{ x }^{ 2 } } \right)$

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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