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CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 1 - MCQExams.com
CBSE
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
Explanation
The derivative or the essence of differentiation is the instantaneous rate of change of a function with respect to one of its variables.
If $$\displaystyle\mathrm{x}=\mathrm{e}^{\mathrm{y}+\mathrm{e}^{\mathrm{y}+\mathrm{e^y+...\infty}}},\ \mathrm{x}>0$$, 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.
If $$f(1)=1, f'(1)=3$$, 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 $$\mathrm{g}(\mathrm{0})=0$$.
If $$\mathrm{f}(-10\sqrt{2})=2\sqrt{2}$$, 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, $$f(x)=\sqrt{3}\sin x-\cos x-2ax+b$$
$$f'(x)=\sqrt{3}\cos x + \sin x -2a$$
$$f'(x)=2\left (\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)-2a$$
$$f'(x)=2\left (\sin\left (\dfrac{\pi}{3}+x\right)-a\right)$$
if $$a>1$$, 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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0%
True
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False
Explanation
For uniform motion, instantaneous speed is constant. To understand it in simple words we can say that instantaneous speed at any given time is the magnitude of instantaneous velocity at that time. It is a limit of the average
as the time interval becomes very small.
$$\displaystyle \frac{d}{dx}(\tan^{-1}\frac{\sqrt{x}-x}{1+x^{3/2}}.)$$
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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 $$\displaystyle y = \tan^{-1}\sqrt{\left ( \frac{1-\cos x}{1+\cos x} \right )}$$
$$\Rightarrow \displaystyle y = \tan^{-1}\sqrt{\left ( \frac{2\sin^2\frac{x}{2} }{2\cos^2 \frac{x}{2}} \right )}$$
$$\Rightarrow \displaystyle y= \tan^{-1}\left [ \tan\left ( \frac{x}{2} \right ) \right ]=\frac{x}{2}$$
$$\displaystyle \therefore \frac{dy}{dx}= \frac{1}{2}$$
If $$y=2 \sin x -3x^4 + 8$$, 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
If $$y=2 \sin x -3x^4 + 8$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-3(4x^{(4-1)})+0$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-12x^3$$
Therefore, the correct answer is $$B$$.
If $$\displaystyle f(x)=|\cos x|$$ 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
$$\therefore y = - \cos x $$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\sin x$$
$$ \Rightarrow \left ( \dfrac{dy}{dx} \right )_{3\pi /4}=\sin \dfrac{3\pi }{4}= \dfrac{1}{\sqrt{2}}$$
If $$\displaystyle xe^{xy}-y=\sin x $$ then $$\displaystyle \frac{dy}{dx}$$ at $$x = 0$$ is
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$$0$$
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$$1$$
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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)
$$\displaystyle \left ( xy_{1}-y \right )^{2}+x^{2}y^{2}=x^{4}.1=x^4$$
If $$f(x) = \displaystyle \log \left | 2x \right |, x\neq 0 $$ 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 $$x:$$
$$\displaystyle e^{x}x^{5}$$
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$$\displaystyle 5e^{x}x^{4}+e^{x}x^{5}$$
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$$\displaystyle 4e^{x}x^{5}+e^{x}x^{5}$$
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$$\displaystyle 5e^{x}x^{4}+e^{x}x^{4}$$
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$$\displaystyle 4e^{x}x^{5}+e^{x}x^{4}$$
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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$$3x^{2}-2$$
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$$3x^{2}+10x-2$$
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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$$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=a$$
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$$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=-a$$
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$$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=2a$$
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$$\displaystyle x\left ( \frac{dy}{dx} \right )^{2}=-2a$$
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-
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$$\displaystyle e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$$
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$$\displaystyle 2e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$$
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$$\displaystyle \frac{1}{2}e^{x}\sin (\sqrt{3}x+\dfrac{\pi}3 )$$
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$$\displaystyle \frac{1}{2}e^{x}\sin (\sqrt{3}x-\dfrac{\pi}3 )$$
Explanation
Let $$\displaystyle y=e^{x}\sin \sqrt{3}\: \: x$$
$$\therefore \displaystyle \frac{dy}{dx}=e^{x}\sin \sqrt{3} x+e^x\sqrt{3}\cos \sqrt{3}x$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=2e^{x}\left [ \frac{1}{2}\sin \sqrt{3x}+\frac{\sqrt{3}}{2}\cos \sqrt{3}x \right ]$$
$$\displaystyle \Rightarrow \frac{dy}{dx}=2.e^{x}\sin \left ( \sqrt{3}x+\frac{\pi }{3} \right )$$
If $$\displaystyle x+y=x^{y}$$ then $$\displaystyle \frac{dy}{dx}\ equals-$$
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$$\displaystyle \frac{yx^{y-1}-1}{1-x^{y}\log x}$$
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$$\displaystyle \frac{yx^{y-1}-1}{x^{y}\log x-1}$$
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$$\displaystyle \frac{yx^{y-1}+1}{x^{y}\log x+1}$$
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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
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$$yx^{y - 1}$$
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$$\displaystyle \frac{y^2}{x(1 - y log x)}$$
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$$\displaystyle \frac{y}{x(1 + y log x)}$$
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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 $$x $$
$$\displaystyle \frac{1}{y}. \frac{dy}{dx} = y . \frac{1}{x} + \frac{dy}{dx}. log x$$
$$\therefore \displaystyle \frac{dy}{dx} = \frac{y}{x} . \frac{y}{(1 - y log x)} = \frac{y^2}{x (1 - y. log x)}$$
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