MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 1

Differentiation gives us the instantaneous rate of change of one variable with respect to another.

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True
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False

$\displaystyle\mathrm{x}=\mathrm{e}^{\mathrm{y}+\mathrm{e}^{\mathrm{y}+\mathrm{e^y+...\infty}}},\ \mathrm{x}>0$ , then

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ is

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$\displaystyle \frac{\mathrm{x}}{1+\mathrm{x}}$
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$\displaystyle \frac{1}{\mathrm{x}}$
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$\displaystyle \frac{1-\mathrm{x}}{\mathrm{x}}$
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$\displaystyle \frac{1+\mathrm{x}}{\mathrm{x}}$

Explanation

Given,

$\displaystyle x={ e }^{ y+{ e }^{ y+... } }$

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

Taking

$\log$ both sides

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

Differentiating both sides w.r.t

$x$ , we get

$\Rightarrow \displaystyle \frac { 1 }{ x } =\frac { dy }{ dx } +1$

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { 1-x }{ x }$

Let

$\mathrm{y}$ be an implicit function of

$\mathrm{x}$ defined by

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

$\mathrm{y}'(1)$ equals

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$-1$
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$1$
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$\log 2$
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$-\log 2$

Explanation

When

$\mathrm{x}= 1,\ \mathrm{y} =\displaystyle \frac{\pi}{2}$

$(\mathrm{x}^{\mathrm{x}}-\cot \mathrm{y})^{2}=[cosec]^{2}\mathrm{y}$

$\mathrm{x}^{\mathrm{x}}= coty + |cosec y|$

when

$\mathrm{x}=1,\ \displaystyle \mathrm{y}=\frac{\pi}{2}$

$\Rightarrow \mathrm{x}^{\mathrm{x}}=$ coty

$+$ cosec

$\mathrm{y}$

diff. w.r.t.

$\mathrm{x}$

$\mathrm{x}^{\mathrm{x}}(1+1\mathrm{n}\mathrm{x})=$ (

$-[cosec]^{2}\mathrm{y}$

$-cosecy\cot \mathrm{y}$ )

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

when

$\mathrm{x}=1$ and

$\displaystyle \mathrm{y}=\frac{\pi}{2}$

$\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-1$
Hence, option 'A' is correct.

$f(1)=1, f'(1)=3$ , then the value of derivative of

$f(f(fx)))+(f(x))^2$ at

$x=1$ is?

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$9$
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$33$
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$12$
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$20$

Explanation

Let

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

Differentiating both sides w.r.t.x

$\dfrac{dy}{dx}=f'(f(f(x))\cdot f'(f(x))\cdot f'(x)+2f(x)f'(x)$
Put

$x=1$

$\Rightarrow f'(f(f(1)))\cdot f'(f(1))\cdot f'(1)+2f(1)\cdot f'(1)$

$=3\times 3\times3+2\times3$ given

$f(1)=1 \text{and} f^{\prime}(1)=3$

=27+6=33

Hence, Option (B) is correct.

For

$x\in R, f(x) = |\log2 - \sin x|$ and

$g(x) = f(f(x))$ , then:

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$g$ is not differentiable at $x = 0$
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$g'(0) = \cos (\log 2)$
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$g'(0) = -\cos (\log 2)$
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$g$ is differentiable at $x = 0$ and $g'(0) = -\sin (\log 2)$

Explanation

$g(x)\quad =f(f(x))$

$g^{'}(x)\quad =f'(f(x))f'(x)$

$g^{'}(0) =f'(f(0))f'(0)$

$x$ tending to

$0$ ,

$log2>sinx$

$f(x)=log2-sinx$

$f'(x) =-cosx$

$f'(0)=-1$

$f'(log 2)=-cos(log 2)$

$g'(0)=(-cos(log 2))(-1) =cos(log 2)$

Consider the functions defined implicitly by the equation

$\mathrm{y}^{3}-3\mathrm{y}+\mathrm{x}=0$ on various intervals in the real line. If

$x \in(-\infty, -2)\cup (2, \infty)$ , the equation implicitly defines a unique real valued differentiable function

$\mathrm{y}=\mathrm{f}(\mathrm{x})$ . If

$x \in(-2,2)$ , the equation implicitly defines a unique real valued differentiable function

$\mathrm{y}=\mathrm{g}(\mathrm{x})$ satisfying

$\mathrm{g}(\mathrm{0})=0$ .

$\mathrm{f}(-10\sqrt{2})=2\sqrt{2}$ , then

$\mathrm{f''}(-10\sqrt{2})=$

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$\displaystyle \frac{4\sqrt{2}}{7^{3}3^{2}}$
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$-\displaystyle \frac{4\sqrt{2}}{7^{3}3^{2}}$
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$\displaystyle \frac{4\sqrt{2}}{7^{3}3}$
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$-\displaystyle \frac{4\sqrt{2}}{7^{3}3}$

Explanation

Differentiating the given equation, we get

$3\mathrm{y}^{2}\mathrm{y}'-3\mathrm{y}'+1=0$

$\displaystyle \Rightarrow \mathrm{y}'(-10\sqrt{2})=-\frac{1}{21}$

Differentiation again we get

$6\mathrm{y}\mathrm{y}^{\prime 2}+3\mathrm{y}^{2}\mathrm{y}''-3\mathrm{y}''=0$

$\displaystyle \Rightarrow \mathrm{f}''(-10\sqrt{2})=-\frac{6.2\sqrt{2}}{(21)^{4}}=-\frac{4\sqrt{2}}{7^{3}3^{2}}$

Consider the functions defined implicitly by the equation

$\mathrm{y}^{3}-3\mathrm{y}+\mathrm{x}=0$ on various intervals in the real line. If

$x \in(-\infty, -2)\cup (2, \infty)$ , the equation implicitly defines a unique real valued differentiable function

$\mathrm{y}=\mathrm{f}(\mathrm{x})$ . If

$x \in(-2,2)$ , the equation implicitly defines a unique real valued differentiable function

$\mathrm{y}=\mathrm{g}(\mathrm{x})$ satisfying

$\mathrm{g}(\mathrm{0})=0$ .

$\displaystyle \int_{-\mathrm{l}}^{1}\mathrm{g}'(\mathrm{x}) dx =$

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$2\mathrm{g}(-1)$
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$0$
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$-2\mathrm{g}(1)$
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$2\mathrm{g}(1)$

Explanation

For

$x$ lies in

$(-2,2)$

$(g(x))^3-g(x)+x=0$

$(g(-x))^3-g(-x)-x=0$

Adding we get

$[g(x)+g(-x)][g(x)^2+g(-x)^2-g(x)g(-x)-3]=0$

Suppose

${g(x)}^2+{g(-x)}^2-g(x)g(-x)-3=0$

Put

$\ x=0$ . Then we get

${g(0)}^2+{g(0)}^2-g(0)g(0)-3$

$0-3=0$ , Not possible.

Thus,

$g(x)+g(-x)=0$

Hence

$\displaystyle \int_{-\mathrm{1}}^{1}\mathrm{g}'(\mathrm{x})dx=\mathrm{g}(1)-\mathrm{g}(-1)=2\mathrm{g}(1)$

The value of '

$a$ ' in order

$f(x)=\sqrt{3}\sin x-\cos x -2ax+b$ decrease for all real values of

$x$ , is given by

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$a>1$
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$a \ge 1$
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$a \ge \overline{2}$
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$a < \overline{2}$

Explanation

Given,

$f(x)=\sqrt{3}\sin x-\cos x-2ax+b$

$f'(x)=\sqrt{3}\cos x + \sin x -2a$

$f'(x)=2\left (\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)-2a$

$f'(x)=2\left (\sin\left (\dfrac{\pi}{3}+x\right)-a\right)$

$a>1$ , then for all real values of

$x$

$,f'(x)<0$ .

So,

$a>1$

$\displaystyle \frac{d}{dx}(xe^{x})$

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$xe^{x}+e^{x}$
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$xe^{x}-e^{x}$
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$xe^{x}+e^{2x}$
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$xe^{x}-e^{2x}$

Explanation

Let

$y =\displaystyle xe^{x}$

differentiating w.r.t x,

$\displaystyle \frac{dy}{dx}= \lim_{\delta x\rightarrow 0}\frac{\left ( x+\delta x \right )e^{x+\delta x}-xe^{x}}{\delta x}$

$\displaystyle= \lim_{\delta x\rightarrow 0}\frac{x\left [ e^{x+\delta x}-e^{x} \right ]}{\delta x}+\frac{\delta x.e^{x+\delta x}}{\delta x}= xe^{x}+e^{x}$

$\displaystyle \frac{d}{dx}[f(x)\cdot g(x)] =f(x) \frac{d}{dx}g(x)+g(x) \frac{d}{dx}f(x)$ is known as _____ rule.

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Product
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Sum
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Multiplication
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None of these

Explanation

The product rule for two functions

$f(x)$ and

$g(x)$ is given by

$\displaystyle \frac{d}{dx}[f(x)\cdot g(x)] =f(x) \frac{d}{dx}g(x)+g(x) \frac{d}{dx}f(x)$

For instantaneous speed, the distance traveled by the object and the time taken are both equal to zero.

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True
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False

$\displaystyle \frac{d}{dx}(\tan^{-1}\frac{\sqrt{x}-x}{1+x^{3/2}}.)$

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$\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$
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$\displaystyle \frac{1}{1-x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$
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$\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{3}}.$
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$-\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$

Explanation

Let

$\displaystyle y=\tan^{-1}\frac{\sqrt{x}-x}{1+\sqrt{\left ( x \right )}.x}$

$=\tan^{-1}\sqrt{x}-\tan^{-1}x$

$\displaystyle \therefore \frac{dy}{dx}=\frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}[l \mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{x})^{\mathrm{x}}]$ , where

$a$ is a constant, is equal to

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$1$
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$\log {ax}$
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$1/a$
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$\log {(ax)+1}$

Explanation

$\frac{d}{dx}\left ( x \log ax \right )= \log (ax) +\dfrac{ax}{ax}=1+\log(ax)$

Derivative of

$2\tan x - 7\sec x$ with respect to

$x$ is:

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$2 \sec x + 7 \tan x$
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$\sec x (2 \sec x + \tan x)$
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$2 {\sec}^2 x + \sec x. \tan x$
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$\sec x (2 \sec x - 7 \tan x)$

Explanation

Diffrentiating the given function with respect to

$x$ :

$\displaystyle \frac{d}{dx} (2 \tan x - 7 \sec x) = 2 \sec^2 x - 7 \sec x \tan x$

$= \sec x (2 \sec x - 7 \tan x)$

$\displaystyle \frac{d}{dx}( x^{2}e^{ax})$

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$\displaystyle e^{ax}\left ( ax^{2}+2x \right )$
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$\displaystyle e^{ax}\left ( 2ax^{2}+2x \right )$
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$\displaystyle e^{ax}\left ( ax^{2}+2ax \right )$
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$\displaystyle e^{ax}\left ( ax^{2}-2ax \right )$

Explanation

Let

$y = x^2e^{ax}$
Thus using product rule,

$\displaystyle \frac{dy}{dx}=x^2(ae^{ax})+e^{ax}(2x)=e^{ax}(ax^2+2x)$

$\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\cos x-\sin x}{\cos x+\sin x})$

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$-1$
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$-2$
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$1$
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$x$

Explanation

Let

$\displaystyle y = \tan ^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}=\tan ^{-1}\frac{1-\tan x}{1+\tan x}$

$\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}-x)=\frac{\pi}{4}-x$

$\therefore \cfrac{dy}{dx}=-1$

$\displaystyle \frac{d}{dx}(\tan^{-1}\sqrt{\left ( \frac{1-\cos x}{1+\cos x} \right )})$

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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{4}$
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$\displaystyle \frac{1}{\sqrt2}$
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$\displaystyle \frac{-1}{2}$

Explanation

Let

$\displaystyle y = \tan^{-1}\sqrt{\left ( \frac{1-\cos x}{1+\cos x} \right )}$

$\Rightarrow \displaystyle y = \tan^{-1}\sqrt{\left ( \frac{2\sin^2\frac{x}{2} }{2\cos^2 \frac{x}{2}} \right )}$

$\Rightarrow \displaystyle y= \tan^{-1}\left [ \tan\left ( \frac{x}{2} \right ) \right ]=\frac{x}{2}$

$\displaystyle \therefore \frac{dy}{dx}= \frac{1}{2}$

$y=2 \sin x -3x^4 + 8$ , then

$\dfrac{dy}{dx}$ is

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$2\sin x -12 x^3$
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$2 \cos x-12 x^3$
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$2 \cos x+12 x^3$
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$2 \sin x+12 x^3$

Explanation

$y=2 \sin x -3x^4 + 8$

$\Rightarrow \cfrac{dy}{dx}=2 \cos x-3(4x^{(4-1)})+0$

$\Rightarrow \cfrac{dy}{dx}=2 \cos x-12x^3$
Therefore, the correct answer is

$B$ .

$\displaystyle f(x)=|\cos x|$ then

$f'\left ( \frac{3\pi }{4} \right )$ is equal to-

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$\displaystyle -\frac{1}{\sqrt{2}}$
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$\displaystyle \frac{1}{\sqrt{2}}$
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$1$
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None of these

Explanation

Clearly,

$\cfrac{\pi}{2}<\cfrac{3\pi}{4} < \pi \Rightarrow \cos x< 0$ in this interval

$\therefore y = - \cos x$

$\Rightarrow \displaystyle \frac{dy}{dx}=\sin x$

$\Rightarrow \left ( \dfrac{dy}{dx} \right )_{3\pi /4}=\sin \dfrac{3\pi }{4}= \dfrac{1}{\sqrt{2}}$

$\displaystyle xe^{xy}-y=\sin x$ then

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

$x = 0$ is

0%
$0$
0%
$1$
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$-1$
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None of these

Explanation

Putting

$x=0$ we get,

$y=0$
Now,

$\displaystyle xe^{xy}-y=\sin x$
Differentiating w.r.t

$x$

$\displaystyle e^{xy}+x.e^{xy}\left \{ y+x.\frac{dy}{dx} \right \}-\frac{dy}{dx}=\cos x$
at

$x = 0, y = 0$

$\displaystyle e^{0}+0.e^{0}({0+0})-\frac{dy}{dx}=\cos 0$

$\Rightarrow \displaystyle -\frac{dy}{dx}=0$

Differentiate with respect to x

$\displaystyle x^{4}+3x^{2}-2x$

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$\displaystyle 4x^{3}+6x-2$
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$\displaystyle 4x^{3}+6x-3$
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$\displaystyle 4x^{4}+6x-2$
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None of the above

Explanation

Given,

$y=\displaystyle x^{4}+3x^{2}-2x$

Now, differentiating w.r.t

$x$ , we get

$\dfrac{dy}{dx}=\dfrac{d(x^4)}{dx} + \dfrac{d(3x^2)}{dx}-\dfrac{d(2x)}{dx}$

$\dfrac{dy}{dx}=4{ x }^{ 3 }+6x-2$

$\displaystyle \frac{d}{dx}(x^{2}\cos x)$

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$\displaystyle -x^{2}\sin x+2x\cos x.$
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$\displaystyle -x^{2}\sin x-2x\cos x.$
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$\displaystyle x^{2}\sin x+2x\cos x.$
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$\displaystyle x^{2}\sin x-2x\cos x.$

Explanation

Let

$y = x^2\cos x$
Thus using product rule,

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

Obtain the differential equation whose solution is

$\displaystyle y=x\sin \left ( x+A \right ),$ A being constant.

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

Explanation

Given,

$\displaystyle y=x\sin \left ( x+A \right )$

Differentiating, we get

$\displaystyle y_{1}=\sin \left ( x+A \right )+x\cos \left ( x+A \right)$

$\displaystyle \therefore xy_{1}-y=x^{2}\cos \left ( x+A \right ) ..(1)$

Also

$\displaystyle xy=x^{2}\sin \left ( x+A \right ). ..(2)$

Squaring and adding (1) and (2)

$\displaystyle \left ( xy_{1}-y \right )^{2}+x^{2}y^{2}=x^{4}.1=x^4$

$f(x) = \displaystyle \log \left | 2x \right |, x\neq 0$ then

$f'(x)$ is equal to-

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$\displaystyle \frac{1}{x}$
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$\displaystyle -\frac{1}{x}$
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$\displaystyle \frac{1}{\left | x \right |}$
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None of these

Explanation

$f(x) = \displaystyle \log \left | 2x \right |$

$\displaystyle x\neq 0$

$\displaystyle \log \left | x \right |$ is defined for

$| x | > 0$

$f(x) = \displaystyle \log 2x = \log 2+\log x$

$\displaystyle f'\left ( x \right )=0+\frac{1}{x}=\frac{1}{x}$

Differentiate with respect to

$x:$

$\displaystyle e^{x}x^{5}$

0%
$\displaystyle 5e^{x}x^{4}+e^{x}x^{5}$
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$\displaystyle 4e^{x}x^{5}+e^{x}x^{5}$
0%
$\displaystyle 5e^{x}x^{4}+e^{x}x^{4}$
0%
$\displaystyle 4e^{x}x^{5}+e^{x}x^{4}$

Explanation

We have to find the differentiation of

$e^xx^5$

Consider

$u=e^x$ and

$v=x^5$ .

Then,

we know that,

$\dfrac{d(u \ . v)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$

$\dfrac{d (e^x x^5)}{dx} = x^5 \dfrac{de^x}{dx} + e^x\dfrac{dx^5}{dx}$

$\dfrac{d (e^x x^5)}{dx} = e^xx^5 + 5e^xx^4$ .

Differentiation of

$\displaystyle x^{3}+5x^{2}-2$ with respect to

$x$ is

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$3x^{2}+10x$
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$3x^{2}+10$
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$3x^{2}-2$
0%
$3x^{2}+10x-2$

Explanation

Given,

$x^3+5 x^2-2$

Differentiating with respect to

$x$ , we get

$\displaystyle \frac{d}{dx}\left ( x^{3}+5x^{2}-2 \right )=\frac{d}{dx}\left ( x^{3} \right )+5\frac{d}{dx}\left ( x^{2} \right )-\frac{d}{dx}\left ( 2 \right )$

$=\displaystyle 3x^{2}+5\left ( 2x \right )-0$

$=3x^{2}+10x$

Find the differential equations of all parabolas each having latus rectum

$4a$ and whose axes are parallel to the x-axis.

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$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=a$
0%
$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=-a$
0%
$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=2a$
0%
$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=-2a$

Explanation

The equation of the parabola is

$\displaystyle \left ( y-k \right )^{2}=4ax \cdots \left ( 1 \right )$

$\displaystyle \therefore 2\left( y-k \right )y_{1}=4a$
or

$y-k=\cfrac{2a}{y_{1}}$
Putting in (1), we get

$\displaystyle \frac{4a^{2}}{y^{2}_{1}}=4ax$ or

$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=a$

$\displaystyle \frac{d}{dx}(e^{x}\sin \sqrt{3}x)$ equals-

0%
$\displaystyle e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$
0%
$\displaystyle 2e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$
0%
$\displaystyle \frac{1}{2}e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$
0%
$\displaystyle \frac{1}{2}e^{x}\sin (\sqrt{3}x-\dfrac{\pi}3 )$

Explanation

Let

$\displaystyle y=e^{x}\sin \sqrt{3}\: \: x$

$\therefore \displaystyle \frac{dy}{dx}=e^{x}\sin \sqrt{3} x+e^x\sqrt{3}\cos \sqrt{3}x$

$\Rightarrow \displaystyle \frac{dy}{dx}=2e^{x}\left [ \frac{1}{2}\sin \sqrt{3x}+\frac{\sqrt{3}}{2}\cos \sqrt{3}x \right ]$

$\displaystyle \Rightarrow \frac{dy}{dx}=2.e^{x}\sin \left ( \sqrt{3}x+\frac{\pi }{3} \right )$

$\displaystyle x+y=x^{y}$ then

$\displaystyle \frac{dy}{dx}\ equals-$

0%
$\displaystyle \frac{yx^{y-1}-1}{1-x^{y}\log x}$
0%
$\displaystyle \frac{yx^{y-1}-1}{x^{y}\log x-1}$
0%
$\displaystyle \frac{yx^{y-1}+1}{x^{y}\log x+1}$
0%
None of these

Explanation

$x+y=x^y$
Taking

$\log$ both side,

$\log (x+y)=y \log x$
Now differentiating w.r.t

$x$

$\cfrac{1}{x+y}(1+\cfrac{dy}{dx})=\log x\cfrac{dy}{dx}+\cfrac{y}{x}$

$\Rightarrow 1+\cfrac{dy}{dx}=(x+y)\log x \cfrac{dy}{dx}+\cfrac{y(x+y)}{x}=x^y\log x \cfrac{dy}{dx}+yx^{y-1}$

$\Rightarrow \cfrac{dy}{dx}=\displaystyle \frac{yx^{y-1}-1}{1-x^{y}\log x}$

Let

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

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

0%
$yx^{y - 1}$
0%
$\displaystyle \frac{y^2}{x(1 - y log x)}$
0%
$\displaystyle \frac{y}{x(1 + y log x)}$
0%
None of these

Explanation

We have

$y = x^{x^{x .....}}$
As y has repeated infinite powers of

$x$ so it can be written as

$y = x^y$
Taking log of both sides, we get

$\log y = y. \log x$
Differentiating with respect to

$x$

$\displaystyle \frac{1}{y}. \frac{dy}{dx} = y . \frac{1}{x} + \frac{dy}{dx}. log x$

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

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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