MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 1

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

0%
True
0%
False

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

0%
$a>1$
0%
$a \ge 1$
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$a \ge \overline{2}$
0%
$a < \overline{2}$

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,

$a>1$

$\displaystyle \frac{d(\tan x.)}{dx}$

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$\sec^2 x$
0%
$\cot^2 x$
0%
$\cos^2 x$
0%
$\sin^2 x$

Explanation

$\displaystyle y=\tan x.$

$\displaystyle \therefore \dfrac{dy}{dx} = \lim_{\delta x\rightarrow 0}\dfrac{\tan \left ( x+\delta x \right )-\tan x}{\delta x}$
(Change to

$\sin$ and

$\cos$ ).

$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \left ( x+\delta x-x \right )}{\delta x.\cos x\cos \left(x+\delta x \right)}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \delta x}{\delta x}.\dfrac{1}{\cos x\cos \left(x+\delta x \right)}$

$\displaystyle =1. \sec^{2}x= \sec^{2}x.$

$\displaystyle \frac{d(\sin x)}{dx}.$

0%
$\cos x$
0%
$\sec x$
0%
$-\cos x$
0%
$- \tan x$

Explanation

$\displaystyle y= \sin x= f\left ( x \right ),$ (say)

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

$\displaystyle \therefore \delta y = \left ( x+\delta x \right )-\sin x= f\left ( x+\delta x \right )-f (x)$

$\displaystyle \therefore \frac{\delta y}{\delta x}= \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}= \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}$

$\displaystyle \therefore \frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0} \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}$

$\displaystyle = lim_{\delta x\rightarrow 0} = \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}$

$\displaystyle = \lim_{\delta x\rightarrow 0} = \frac{2\cos \left ( x+\frac{1}{2}\delta x \right )\sin\frac{1}{2}\delta x}{\delta x}$

$\displaystyle = \lim_{\delta x\rightarrow 0}\cos \left ( x+\frac{\delta x}{2}\right )\frac{\sin \left ( \delta x/2 \right )}{\left (\delta x/2 \right )}$

$\displaystyle =\cos x.1= \cos x \left ( \because \lim_{\delta x\rightarrow 0} \frac{\sin \theta }{\theta }= 1.\right )$

Differentiate

$\displaystyle \cos x$

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$\displaystyle \cos x$
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$\displaystyle \cos^2 x$
0%
$\displaystyle \sin x$
0%
$\displaystyle -\sin x$

Explanation

$y=\cos x$

$\displaystyle\frac{dy}{dx}=\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(x) }{ h } }$

$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {\cos(x+h)-\cos(x)}{ h } }$

$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {-2\sin\left(\displaystyle\frac{2x+h}{2}\right)\sin(h/2)}{ h } }$

$=-\sin x$

$\therefore \displaystyle\frac{dy}{dx}=-\sin x$

$\displaystyle \dfrac{d}{dx} \sec x=$

0%
$\sec x \tan x$
0%
$\cos x \tan x$
0%
$\sin x \tan x$
0%
$\sec x \cot x$

Explanation

$\displaystyle y= \sec x$

$\displaystyle \therefore \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sec \left ( x+\delta x \right )-\sec x}{\delta x}$ ....(Change to cos)

$\displaystyle = \lim_{\delta x\rightarrow 0}\frac{\cos x- \cos \left ( x+\delta x \right )}{\delta x.\cos\left ( x+\delta x \right )\cos x}$

We have used

$\cos C-\cos D$ , formula

$\displaystyle = \sin x.\frac{1}{\cos x \cos x}= \sec x\tan x$

State if the given statement is True or False

Derivative of

$y= \cos x$ with respect to

$x$ is

$\sin x$ .

0%
True
0%
False

Explanation

$y=cos x$
Then, the derivative of

$y$ with respect to

$x$ i.e.,

$\dfrac{dy}{dx}=-\sin x$ .
Therefore, the given statement is false.

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

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True
0%
False

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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$\sec x (2 \sec x + \tan x)$
0%
$2 {\sec}^2 x + \sec x. \tan x$
0%
$\sec x (2 \sec x - 7 \tan x)$

Explanation

Diffrentiating the given function with respect to

$x$ :

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

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

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

0%
$sin2x$
0%
$cos2x$
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$sin4x$
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$cos4x$

Explanation

$\displaystyle y= \sin ^{2}x$

$\displaystyle \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x \right )-\sin ^{2}x}{\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x +x\right ).\sin \left ( x+\delta x-x \right )}{\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\sin \left ( 2x+\delta x \right )\frac{\sin \delta x}{\delta x}= \sin 2x.$
We have used

$[\displaystyle \sin^2 A-\sin ^{2}B= \sin \left ( A+B \right )\sin \left ( A-B \right )]$

$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$

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1
0%
0
0%
does not exist
0%
$\infty$

Explanation

$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at}$

$x\rightarrow 0$

$but\ have\ finite\ value = 0$

$0\times\ finite\ value\ =0$

$\displaystyle \frac{d}{dx}(\tan ^{2}ax).$

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$2 a\tan ax \sec^{2}ax.$
0%
$-2 a\tan ax \sec^{2}ax.$
0%
$a\tan ax \sec^{2}ax.$
0%
$2 a \cot ax \sec^{2}ax$

Explanation

$\displaystyle y= \tan ^{2}ax.$

$\therefore \displaystyle \frac{dy}{dx}= \lim_{\delta x\rightarrow 0}\frac{\tan ^{2}a\left ( x+\delta x \right )-\tan ^{2}ax}{\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\tan a\left ( x+\delta x \right )-\tan ax}{\delta x}$

$\displaystyle =[\tan a\left ( x+\delta x \right )+\tan ax]$

$\displaystyle =\lim_{\delta x\rightarrow 0}A\frac{\sin \left ( a\delta x \right )}{a\delta x}\cdot \left [ \frac{\tan a\left ( x+\delta x \right )+\tan ax}{\cos a\left ( x+\delta x \right )\cos ax} \right ]$

$\displaystyle = a.1.\frac{2\tan ax}{\cos ^{2}ax}= 2 a\tan ax sec^{2}ax.$

Differentiate with respect to x

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

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$\displaystyle 4x^{3}+6x-2$
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$\displaystyle 4x^{3}+6x-3$
0%
$\displaystyle 4x^{4}+6x-2$
0%
None of the above

Explanation

Given,

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

Now, differentiating w.r.t

$x$ , we get

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

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

$\displaystyle \frac{d}{dx}\left(\frac{\sin x}{x}\right)$

0%
$\dfrac{x\cos x-\sin x}{x^{2}}.$
0%
$\dfrac{x\cos x+\sin x}{x^{2}}.$
0%
$\dfrac{x\cos x+\sin x}{x^{3}}.$
0%
$\dfrac{x\cos x-\sin x}{x^{3}}.$

Explanation

$\displaystyle y=\dfrac{\sin x}{x}$

$\displaystyle \dfrac{dy}{dx}= \lim_{\delta x\rightarrow 0}\dfrac{1}{\delta x}\left [ \dfrac{\sin\left ( x+\delta x \right )}{x+\delta x}-\dfrac{\sin x}{x} \right ]$

$\displaystyle = \lim_{\delta x\rightarrow 0}\dfrac{x\sin\left ( x+\delta x \right )-\left ( x+\delta x \right )\sin x }{x\left ( x+\delta x \right ).\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\left [ \dfrac{1}{x+\delta x}.\dfrac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}-\dfrac{\sin x}{x\left ( x+\delta x \right )} \right ]$

$\displaystyle = \dfrac1x\dfrac{d(\sin x)}{dx} - \dfrac{\sin x}{x^2}$

$\displaystyle = \dfrac{1}{x}.\cos x-\dfrac{\sin x}{x^{2}}= \dfrac{x\cos x-\sin x}{x^{2}}.$

$\displaystyle \frac{d\sin x^{2}}{dx}$

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

Explanation

$\displaystyle y= \sin x^{2}$

$\displaystyle \frac{dy}{dx}= \lim_{\delta x\rightarrow 0}\frac{\sin \left ( x+\delta x \right )^{2}-\sin x^2}{\delta x}$

$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x^{2}+2x\delta x^{2}+(\delta x)^2\right )-\sin x^{2}}{\delta x}$

$=\displaystyle \lim_{\delta x\rightarrow 0}\frac{2\cos \left ( x^{2}+x\delta x+\frac{1}{2}\delta x^{2} \right )\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{\delta x}$

$=\displaystyle 2\cos x^{2}\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{x\delta x+\frac{1}{2}\delta x^{2}}.\frac{x\delta x+\frac{1}{2}\delta x^{2}}{\delta x}$

$\displaystyle =2\cos x^{2}.1.x = 2x\cos x^{2}$

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

0%
$\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$
0%
$\displaystyle \frac{1}{1-x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$
0%
$\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{3}}.$
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$-\displaystyle \frac{1}{1+x}.\frac{1}{2\sqrt{\left ( x \right )}}-\frac{1}{1+x^{2}}.$

Explanation

Let

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

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

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

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

$f'(x)$ is equal to-

0%
$\displaystyle \frac{1}{x}$
0%
$\displaystyle -\frac{1}{x}$
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$\displaystyle \frac{1}{\left | x \right |}$
0%
None of these

Explanation

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

$\displaystyle x\neq 0$

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

$| x | > 0$

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

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

Differentiation of

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

$x$ is

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

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

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

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{\sqrt2}$
0%
$\displaystyle \frac{-1}{2}$

Explanation

Let

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

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

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

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

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

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

Explanation

Let

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

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

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

$\displaystyle \lim_{x\rightarrow \infty} \sin x$ equals

0%
$1$
0%
$0$
0%
$\infty$
0%
does not exist

Explanation

Let

$L = \displaystyle \lim_{x\rightarrow \infty} \sin x$
Assume

$\displaystyle y = \frac{1}{x}$ so as

$x\to \infty, y \to 0$

$\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}$
We know

$\sin x$ lie between -1 to 1
so let

$p =\sin x$ as

$x\to \infty$
Thus left hand limit

$=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p$
and right hand limit

$=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p$
Clearly L.H.L

$\neq$ R.H.L

$\Rightarrow$ Required limit does not exist.

$x$ is very large, then

$\dfrac {2x}{1+x}$ is

0%
close to $0$
0%
arbitrarily large
0%
lie between $2$ and $3$
0%
close to $2$

Explanation

$\dfrac {2x}{1+x}=\dfrac {2}{1+\dfrac {1}{x}}=\dfrac{2}{1+0}=2$

$x\rightarrow \alpha$ then

$\dfrac {1}{x}\rightarrow 0$ .

What is

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } }$ equal to?

0%
$0$
0%
$\pi$
0%
$\dfrac { 1 }{ \pi }$
0%
$1$

Explanation

Here, putting

$x=0$ directly we have,

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } }$

$=\dfrac {\cos 0 }{ \pi -0 } =\dfrac { 1 }{ \pi }$

Hence, C is correct.

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }$ is equal to

0%
$3$
0%
$1$
0%
$0$
0%
$2$

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }$

$=\displaystyle \lim_{x\to 0}\dfrac{xe^x}{x}-\lim_{x\to 0}\dfrac{\sin x}{x}$ , split up

$=\displaystyle \lim_{x\to 0}e^x-\lim_{x\to 0}\dfrac{\sin x}{x}$

$=e^0-1=1-1=0$

Use limit properties to evaluate

$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x$

0%
$12$
0%
$14$
0%
$16$
0%
$18$

Explanation

$\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x =\lim_{x\to 4}\left(3x\tan\dfrac{\pi}{x}\right)$ , simplify

$=3(4)\tan\dfrac{\pi}{4}=12(1)=12$ , [substitute the limit ]

The from of the limit is not indeterminate, so we can substitute the limit value

Evaluate

$\underset{x \rightarrow 3}\lim \sqrt[4] {x^3}$ using the properties of limits.

0%
$28^{1/4}$
0%
$25^{1/4}$
0%
$27^{1/4}$
0%
$26^{1/4}$

Explanation

$\displaystyle \lim _{ x\rightarrow 3 }{ \sqrt [ 4 ]{ { x }^{ 3 } } } \\={ (3^3) }^{ { 1 }/{ 4 } }={ (27) }^{ { 1 }/{ 4 } }$

$\displaystyle{\lim_{x \to 0}}$

$\Bigg(\dfrac{(1+x)^{2}}{e^{x}}\Bigg)^\dfrac{4}{\sin x}$ is:

0%
$e^2$
0%
$e^{4}$
0%
$e^8$
0%
$e^{-8}$

Explanation

Now

$\lim\limits_{x \to 0}\left(\dfrac{(1+x)^2}{e^{x}}\right)^{\dfrac{4}{\sin x}}$

$=\lim\limits_{x \to 0}\dfrac{\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\dfrac{x}{\sin x}}\right)^8}{e^{\cfrac{4x}{\sin x}}}$

We have

$\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$ and

$\lim\limits_{x \to 0}(1+x)^{\dfrac{1}{x}}=e$ .

As both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing, so we can write the above limits as

$=\dfrac{\lim\limits_{x\to 0}\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\lim\limits_{x\to 0}\dfrac{x}{\sin x}}\right)^8}{\lim\limits_{x\to 0}e^{\cfrac{4x}{\sin x}}}$ [Using division property of limits]

$=\dfrac{\left(\lim\limits_{x\to 0}\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\left(\lim\limits_{x\to 0}\dfrac{x}{\sin x}\right)}\right)^8}{e^{\left(\lim\limits_{x\to 0}\dfrac{4x}{\sin x}\right)}}$ [Using limit property]

$=\dfrac{(e^1)^8}{e^{\frac{4}{1}}}=e^2$

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%
$\cfrac { dy }{ dx } =-\sqrt { x } \left( 3+5x \right)$
0%
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 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= $e^{x^3} +\frac{logx}{x}$

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

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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