If $$\sqrt{x+y}+\sqrt{y-x}=c$$, then $$\dfrac{d^2y}{dx^2}$$ is

$$\displaystyle \frac{2}{c^2}$$

$$\displaystyle -\frac{2}{c^2}$$

If $${ e }^{ y }+xy=e,$$ then $$\displaystyle{ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } }_{ x=0 }^{ }$$ is

$$\displaystyle \frac { 1 }{ { e }^{ 2 } } $$

If $$y=\sqrt{\sin{x}+y}$$, then $$\displaystyle\frac{dy}{dx}$$ is

$$\displaystyle\frac{\cos{x}}{2y-1}$$

$$\displaystyle\frac{\cos{x}}{1-2y}$$

$$\displaystyle\frac{\cos{x}}{2y+1}$$

$$\displaystyle\frac{\sin{x}}{2y-1}$$

If $$x^3+y^3=3axy$$, then $$\displaystyle\frac{dy}{dx}$$ is

$$\displaystyle\frac{ay-y^2}{x^2-ax}$$

$$\displaystyle\frac{by-x^2}{y^2-bx}$$

If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.

$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

If $$y=x^2+sin^{-1}x+log_ex$$, find $$\dfrac {dy}{dx}$$

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$

$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 3

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

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

0%
$\displaystyle \frac{1}{x+\sqrt{1+x^{2}}}$
0%
$\displaystyle \frac{x}{\log(x+\sqrt{1+x^{2}})}$
0%
$\displaystyle \frac{-1}{\log(x+\sqrt{1+x^{2}})}$
0%
$\displaystyle \frac{1}{\sqrt{1+x^{2}}\log(x+\sqrt{1+x^{2}})}$

Explanation

We have,

$y =\log(\log(x+\sqrt{1+x^2}))$
Using chain rule of differentiation,

$\displaystyle \frac{dy}{dx}=\frac{1}{\log(x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times \left( 1+\frac{2x}{2\sqrt{1+x^2}}\right)$

$\displaystyle =\frac{1}{\log(x+\sqrt{1+x^2})}\times \frac{1}{(x+\sqrt{1+x^2})}\times \left(\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^2}}\right)$

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

Hence,

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

$\mathrm{a}^{\mathrm{y}}=\log_{\mathrm{a}}(\mathrm{x}^{2}+\mathrm{x}+1)$ , then

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

0%
$\displaystyle \frac{\log_{a}e.(2x+1)}{(x^{2}+x+1)\log(x^{2}+x+1)}$
0%
$\displaystyle \frac{(2x+1)}{(x^{2}+x+1)\log(x^{2}+x+1)}$
0%
$\displaystyle \frac{1}{(x^{2}+x+1)\log(x^{2}+x+1)}$
0%
$\displaystyle \frac{-1}{(x^{2}+x+1)\log(x^{2}+x+1)}$

Explanation

We have,

$a^y =\log_a(x^2+x+1) \Rightarrow (1)$
Now differentiating both sides, w.r.t

$x$

$\Rightarrow a^y(\log a)\dfrac{dy}{dx} =\log_ae \cdot \dfrac{2x+1}{x^2+x+1}$

$\Rightarrow \dfrac{dy}{dx} =\dfrac{\log_ae\cdot (2x+1)}{(x^2+x+1)\cdot \log(x^2+x+1)}$ , using (1)

$\sqrt{x+y}+\sqrt{y-x}=c$ , then

$\dfrac{d^2y}{dx^2}$ is

0%
$\displaystyle \frac{2}{c}$
0%
$\displaystyle -\frac{2}{c^2}$
0%
$\displaystyle \frac{2}{c^2}$
0%
$\displaystyle -\frac{2}{c}$

Explanation

$\sqrt{x+y}=c-\sqrt{y-x}$
On squaring we get,

$2x-c^2=-2c\sqrt{y-x}$
Again squaring we get,

$y = \dfrac{(2x-c^2)^2}{4c^2}+x$

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

$y''=\dfrac{8}{4c^2}=\dfrac{2}{c^2}$

Option C

$f(x)=(ax+b)\cos x + (cx+d)\sin x$ and

$f^{'}(x)=x \cos x$ , for all values of

$x\in R$ , then

$a,b,c,d$ are given by

0%
$a = b = c = d$
0%
$0, 1, -1, 0$
0%
$1, 0, -1, 0$
0%
$0, 1, 1, 0$

Explanation

Given,

$f(x) = (ax+b)\cos x + (cx+d)\sin x$
Thus

$f'(x) = (ax+b)(-\sin x) + a\cos x + c\sin x + (cx+d)\cos x$

$= (-ax-b+c)\sin x + (a+cx+d)\cos x$
Also given,

$f'(x) = x\cos x$

Comparing the coefficients, we get

$a =0, b=1, c =1, d=0$

The set onto which the derivative of the function

$f(x)=x(\log x -1)$ maps the ray

$[1, \infty)$ is ?

0%
$[1, \infty)$
0%
$(10, \infty)$
0%
$[0,\infty)$
0%
$(0,0)$

Explanation

Given function

$f\left( x \right)=x\left( \log { x } -1 \right)$

differentiate w.r.t

$x$

we get,

$f^{ \prime }\left( x \right) =\cfrac { x }{ x } +\log { x } -1\\ f^{ \prime }\left( x \right) =\log { x }$

Clearly,

$\log x$ is defined in the given region.

$x\rightarrow 1,\log { x } =0\\ x\rightarrow \infty ,\log { x } =\infty$

The set onto which the derivative of the function maps is [

$0,\infty$ ).

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

$(1+x^{2})y_{1}+xy=$

0%
$y$
0%
$-y$
0%
$0$
0%
$1$

Explanation

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

Differentiate both sides:

$y_1\sqrt{1+x^{2}}+\dfrac{yx}{\sqrt{1+x^{2}}}=\dfrac{1+\dfrac{x}{\sqrt{1+x^{2}}}}{x+\sqrt{1+x^{2}}}$

$\Rightarrow y_1\sqrt{1+x^{2}}+\dfrac{yx}{\sqrt{1+x^{2}}}=\dfrac{1}{\sqrt{1+x^{2}}}$

$\Rightarrow y_1(1+x^{2})+yx=1$

$f(x)=\displaystyle \frac{x^{2}}{x+a}$ then

$f^{'}(a)=$

0%
4
0%
$\displaystyle \frac{3}{8}$
0%
$\displaystyle \frac{3}{4}$
0%
8

Explanation

Given,

$f(x)=\displaystyle \frac{x^{2}}{x+a}$
using division rule,

$f'(x)=\displaystyle \frac{(x+a)(2x)-x^2(1)}{(x+a)^2}$

$=\dfrac{x(2x+2a-x)}{(x+a)^2}$

$=\dfrac{x(x+2a)}{(x+a)^2}$

$\therefore f'(a)=\displaystyle \frac{a(a+2a)}{(a+a)^2}=\frac{3}{4}$

${ e }^{ y }+xy=e,$ then

$\displaystyle{ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } }_{ x=0 }^{ }$ is

0%
$\displaystyle \frac { 1 }{ { e }^{ 2 } }$
0%
${e}^{-1}$
0%
$e$
0%
None of these

Explanation

We have

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

Differentiating w.r.t

$x$ , we get

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

Differentiating again w.r.t

$x$ , we get

$\displaystyle { e }^{ y }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +e^y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+2\frac { dy }{ dx } +x\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =0$ -------(2)

Put

$x=0$ in

${ e }^{ y }+xy=e$ , we get

$y=1$

Putting

$x=0,y=1$ in (1) we get

$\displaystyle e\frac { dy }{ dx } +1=0\Rightarrow \frac { dy }{ dx } =-\frac { 1 }{ e }$

Putting

$x=0,y=1,\displaystyle \frac { dy }{ dx } =-\frac { 1 }{ e }$ in (2) we get

$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { 1 }{ { e }^{ 2 } }$

$x^{4}+y^{4}-a^{2}xy=0$ defines

${y}$ implicitly as function of

$x$ , then

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

0%
$\displaystyle \frac{4x^{3}-a^{2}y}{4y^{3}-a^{2}x}$
0%
$-\left(\displaystyle \frac{4x^{3}-a^{2}y}{4y^{3}-a^{2}x}\right)$
0%
$\displaystyle \frac{4x^{3}}{4y^{3}-a^{2}x}$
0%
$\displaystyle \frac{-4x^{3}}{4y^{3}-a^{2}x}$

Explanation

Given,

$x^{4}+y^{4}-a^{2}xy=0$

$x^4+y^4=a^2xy$
Differentiate both sides w.r.t

$x$ we get,

$4x^3+4y^3\dfrac{dy}{dx}$

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

$\therefore \dfrac{dy}{dx}=\dfrac{a^2y-4x^3}{4y^3-a^2x}$

$=-\left(\dfrac{4x^3-a^2y}{4y^3-a^2x} \right)$

$\mathrm{y}=(\mathrm{x}^{2}+1)^{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}$ , then

$\mathrm{y}^{'}(0)$ is equal to

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

Explanation

$\mathrm {y'(x)}=\left(\mathrm{x}^{2}+1)^{\sin \mathrm{x}}(\cos \mathrm{x } log(1+\mathrm {x^{2}})+\sin \mathrm { x}\left ( \dfrac{2{\mathrm{x}}}{1+\mathrm{x^{2}}} \right ) \right)$

$\mathrm{y'(0)}=1(0+0)$

$=0$

$f(x)=e$

$^{x}$

$g(x),$

$g(0)=1,g'(0)=3$ , then

$f' (0)$ is

0%
$0$
0%
$4$
0%
$3$
0%
$7$

Explanation

${f}'(x)=e^{^{x}}g(x)+e^{^{x}}{g}'\left ( x \right )$

For x=0:

$f'(0)= g\left ( 0 \right )+{g}'\left ( 0\right )=4$

$ax^{2}+2hxy+by^{2}+2gx+2fy +c=0$ then

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

0%
$-(\displaystyle \frac{ax+hy+g}{hx+by+f})$
0%
$-(\displaystyle \frac{ax+hy+g}{bx+hy+f})$
0%
$-(\displaystyle \frac{hx+by+f}{ax+hy+g})$
0%
$-(\displaystyle \frac{hx+by+f}{hx+ay+g})$

Explanation

Given:

$ax^{2}+2hxy+by^{2}+2gx+2fy +c=0$

$\displaystyle2ax+2hy+2hx\frac{dy}{dx}+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx}=0$

$\dfrac{-(2ax+2hy+2g)}{2hx+2by+2f}=\dfrac{dy}{dx}$

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

$x\displaystyle \frac{d u}{d x}+y\frac{d u}{d y}=$

0%
$\sin 2u$
0%
$\dfrac {1}{2}\sin 2u$
0%
$\dfrac {1}{3}\sin 2u$
0%
$2\sin 2u$

Explanation

Given,

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

$\Rightarrow \tan u=\dfrac{x^{2}+y^{2}}{(x+y)}$

On differentiating both the sides w.r.t.

$x$ and

$y$ respectively, we get

$\sec^{2}u\dfrac{du}{dx}=\dfrac{2x(x+y)-(x^{2}+y^{2})}{(x+y)^{2}}$ ....(1)

and

$\sec^{2}u\dfrac{du}{dy}=\dfrac{2y(x+y)-(x^{2}+y^{2})}{(x+y)^{2}}$ ....(2)

Multiplying

$x$ with equation (1) and

$y$ to equation (2).

Therefore,

$x\sec^2u\dfrac {du}{dx}=\dfrac {x^3+2x^2y-xy^2}{(x+y)^2}$ ....(3)

and

$y\sec^2u\dfrac {du}{dy}=\dfrac {y^3+2xy^2-x^2y}{(x+y)^2}$ ....(4)

Adding equations (3) and (4), we get

$\sec^{2}u\left ( x\dfrac{du}{dx}+y\dfrac{du}{dy} \right )=\dfrac{x^{3}+2x^{2}y-xy^{2}+y^{3}+2xy^{2}-x^{2}y}{(x+y)^{2}}$

$=\dfrac {x^3+x^2y+y^3+xy^2}{(x+y)^2}$

$=\dfrac {(x+y)(x^2+y^2)}{(x+y)^2}$

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

$=\tan u$

Thus

$x\dfrac{du}{dx}+y\dfrac{du}{dy}=\sin u \cos u$

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

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

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

0%
$\displaystyle\frac{\cos{x}}{2y-1}$
0%
$\displaystyle\frac{\cos{x}}{1-2y}$
0%
$\displaystyle\frac{\cos{x}}{2y+1}$
0%
$\displaystyle\frac{\sin{x}}{2y-1}$

Explanation

We have,

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

On squaring both the sides, we get

$y^2=\sin(x)+y$

Differentiating w.r.t.

$x$ ,

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

$2y\displaystyle \frac { dy }{ dx } -\frac { dy }{ dx } =\cos { x }$

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

$\displaystyle\frac{dy}{dx}=\frac{\cos{x}}{2y-1}$

$\sqrt{x}+\sqrt{y}=4$ , then find

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

$y=1$ .

0%
$3$
0%
$4$
0%
$-3$
0%
$-4$

Explanation

$\sqrt{x}+\sqrt{y}=4$
Differentiating both sides of the given equation w.r.t. y, we get

$\displaystyle\frac{1}{2\sqrt{x}}\frac{dx}{dy}+\frac{1}{2\sqrt{y}}=0$

or

$\displaystyle\frac{dx}{dy}=-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{y}-4}{\sqrt{y}}$ ........................................Since

$\sqrt{x}+\sqrt{y}=4$

or

$\displaystyle{\left[\frac{dx}{dy}\right]}_{\displaystyle y=1}=\frac{1-4}{1}=-3$

$x^3+y^3=3axy$ , then

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

0%
$\displaystyle\frac{ay-y^2}{x^2-ax}$
0%
$\displaystyle\frac{ay-x^2}{y^2-ax}$
0%
$\displaystyle\frac{by-x^2}{y^2-bx}$
0%
None of these

Explanation

Given that

$x^3+y^3=3axy$
We have to find

$\displaystyle \frac { dy }{ dx }$
Differentiate w r to

$x$

$\displaystyle \frac { d }{ dx } \left( { x }^{ 3 }+{ y }^{ 3 }-3axy \right) =0$

$3{ x }^{ 2 }+3{ y }^{ 2 }\displaystyle \frac { dy }{ dx } -3a\left( x\displaystyle \frac { dy }{ dx } +y \right) =0$

$3{ x }^{ 2 }+3{ y }^{ 2 }\displaystyle \frac { dy }{ dx } -3ax\displaystyle \frac { dy }{ dx } -3ay=0$
Dividing throughout by

$3$

$\displaystyle \frac { dy }{ dx } \left( { y }^{ 2 }-ax \right) =ay-{ x }^{ 2 }$

$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { ay-{ x }^{ 2 } }{ { y }^{ 2 }-ax }$

For the function

$f(x) = \displaystyle \frac{x^{100}}{100} + \frac{x^{99}}{99} + ........... + \frac{x^2}{2} + x+1$ ,

$f'(1) =$

0%
$x^{100}$
0%
$100$
0%
$101$
0%
None of these

Explanation

$Given:\cfrac { d }{ dx } (x^{ n })=nx^{ n-1 }\\ \therefore for\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...............+\cfrac { x^{ 2 } }{ 2 } +1\\ f'(x)=\cfrac { 100x^{ 99 } }{ 100 } +99\cfrac { x^{ 98 } }{ 99 } +.............+\cfrac { 2x }{ 2 } +1\\ Here\; f'(1)=1+1+.........to\; 100term=\; 100\\ Hence\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...........+\cfrac { x^{ 2 } }{ 2 } +x+1=100$

$y=x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\cdots}}}$ then

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

0%
$\displaystyle\frac{dy}{dx}=\frac{x}{2x-y}$
0%
$\displaystyle\frac{dy}{dx}=\frac{y}{2y-x}$
0%
$\displaystyle\frac{dy}{dx}=\frac{2y}{2x-y}$
0%
None of these

Explanation

We have

$y=x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\displaystyle\frac{1}{x+\cdots}}}$

$=x+\displaystyle\frac{1}{y}$
or

$y^2=xy+1$
or

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

$[$ Differentiating both sides w.r.t.

$x]$

or

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

or

$\displaystyle\frac{dy}{dx}=\frac{y}{2y-x}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Let

$f(x) = (px+q) \displaystyle \left ( \cfrac{r}{x}+s \right )$
We have,

$\cfrac{d(uv)}{dx} = u'v + uv'$

$\therefore \displaystyle f'(x) = \left [ \cfrac{d}{dx}(px+q) \right ] \left ( \cfrac{r}{x} +s \right ) +(px+q) \cfrac{d}{dx} \left ( \cfrac{r}{x} +s \right)$

$=\displaystyle p\left ( \cfrac{r}{x} + s \right ) +(px+q)\left ( \cfrac{-r}{x^2} \right )$

$= \displaystyle \cfrac{pr}{x}+ps -\cfrac{pr}{x}-\cfrac{qr}{x^2}=ps - \cfrac{qr}{x^2}$

Hence, assertion is incorrect but reason is correct.

$\displaystyle x^2+y^2=t-\frac{1}{t}$ and

$\displaystyle x^4+y^4=t^2+\frac{1}{t^2}$ , then

$\displaystyle x^3y\frac{dy}{dx}=$

0%
$0$
0%
$1$
0%
$-1$
0%
none of these

Explanation

We have

$\displaystyle x^2+y^2=t-\frac{1}{t}$ ..........(1)

$\displaystyle x^4+y^4=t^2+\frac{1}{t^2}$ ................(2)
On squaring equation (1), we get

$\displaystyle {(x^2+y^2)}^2=t^2+\frac{1}{t^2}-2$

${ x }^{ 4 }+2{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 4 }=x^4+y^4-2$ ..............from (2)

$\therefore 2x^2y^2=-2$

$x^2y^2=-1$

$\displaystyle y^2=-\frac{1}{x^2}$
Differentiating both the sides,

$\displaystyle 2y\frac{dy}{dx}=\frac{2}{x^3}$

$\displaystyle x^3y\frac{dy}{dx}=1$

$xy+y^2=\tan x + y$ , then

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

0%
$\displaystyle\frac{\sec^{2}{x}-y}{(x+2y-1)}$
0%
$\displaystyle\frac{\cos^{2}{x}+y}{(x+2y-1)}$
0%
$\displaystyle\frac{\sec^{2}{x}-y}{(2x+y-1)}$
0%
$\displaystyle\frac{\cos^{2}{x}+y}{(2x+2y-1)}$

Explanation

The given relation is

$xy+y^2=\tan x + y$ .
Differentiating both sides with respect to

$x$ , we get

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

or

$\displaystyle\left[y.1+x.\frac{dy}{dx}\right]+2y\frac{dy}{dx}=\sec^{2}{x}+\frac{dy}{dx}$

or

$\displaystyle(x+2y-1)\frac{dy}{dx}=\sec^{2}{x}-y$

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

$y^x=x^y$ , then find

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

0%
$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$
0%
$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$
0%
${y(x\log{y}-y)}$
0%
$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$

Explanation

$y^x=x^y$

Taking log on both sides

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

$\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}$

$\Rightarrow$

$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$

Differentiate

$\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ with respect to

$x$ .

0%
$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}-\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$
0%
$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-2}+\frac{1}{x-3}+\frac{1}{x-4}+\frac{1}{x-5}\right]$
0%
$\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$
0%
None of these

Explanation

Let

$y=\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
Taking logarithm on both sides, we get

$\log{y}=log{\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}}$

$\displaystyle =\frac{1}{2}[\log{(x-1)}+\log{(x-2)}-\log{(x-3)}-\log{(x-4)}-\log{(x-5)}]$ ....................(Since

$\log { \displaystyle \frac { a }{ b } =\log { a } } -\log { b }$ and

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

Differentiating both sides w.r.t.

$x$ , we get

$\displaystyle\frac{1}{y}\frac{dy}{dx}=\frac { 1 }{ 2 } \left( \frac { 1 }{ x-1 } \frac { d }{ dx } \left( x-1 \right) +\frac { 1 }{ x-2 } \frac { d }{ dx } \left( x-2 \right) -\frac { 1 }{ x-3 } \frac { d }{ dx } \left( x-3 \right) -\frac { 1 }{ x-4 } \frac { d }{ dx } \left( x-4 \right) -\frac { 1 }{ x-5 } \frac { d }{ dx } \left( x-5 \right) \right)$

$\displaystyle\therefore\frac{dy}{dx}=\frac{1}{2}\sqrt{\displaystyle\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$

$y=x^{-\tfrac12}+\log_5x+\displaystyle \frac {\sin x}{\cos x}+2^x$ , then find

$\dfrac {dy}{dx}$

0%
$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$
0%
$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$

Explanation

Here, we have function

$y=x^{-1/2}+\log _5x+\tan x+2^x$

On differentiating w.r.t x, we get

$\dfrac {dy}{dx}=\dfrac {d}{dx}(x)^{-1/2}+\dfrac {d}{dx}(\log _5x)+\dfrac {d}{dx} \tan x+\dfrac {d}{dx}(2^x)$

$=-\dfrac {1}{2}(x)^{-1/2-1}+\dfrac {1}{x \log _e5}+\sec ^2x+2^x \log 2$

$=-\dfrac {1}{2}x^{-3/2}+\dfrac {1}{x\log _e5}+\sec ^2x+2^x\log 2$

$y=e^x \sin x$ , then find

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

0%
$e^x(\sin x+\cos x)$
0%
$e^x(\sin x-\cos x)$
0%
$e^x \sin x$
0%
None of these

Explanation

Given that

$y={ e }^{ x }.\sin { x }$
On differentiating, we get

$\displaystyle \frac { dy }{ dx } =\sin { x } \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \sin { x }$

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

$\therefore \displaystyle \frac { dy }{ dx } ={ e }^{ x }\left( \sin { x } +\cos { x } \right)$

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

$y=x^x$ is

0%
$x^x(1-\log{x})$
0%
$x^x(1-\log{y})$
0%
$x^x(1+\log{y})$
0%
$x^x(1+\log{x})$

Explanation

$y=x^x$
Taking

$\log$ on both the sides

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

$\log { y } =x\log { x }$
Differentiate w.r to

$x$

$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right)$

$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right]$

$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right]$

$y=x^2+sin^{-1}x+log_ex$ , find

$\dfrac {dy}{dx}$

0%
$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$
0%
$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$
0%
$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$
0%
$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$

Explanation

$y=x^2+sin^{-1}+log_ex$
On differentiating, we get

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

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

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

$y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+\cdots\:to\:\infty}}}$ , then

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

0%
$\displaystyle\frac{\cos{x}}{1+2y}$
0%
$\displaystyle -\frac{\sin{x}}{1-2y}$
0%
$\displaystyle\frac{\cos{x}}{1-2y}$
0%
$\displaystyle\frac{\cos{x}}{2y-1}$

Explanation

We have

$y=\sqrt{\sin{x}+\sqrt{\sin{x}+\sqrt{\sin{x}+\cdots\:to\:\infty}}}$
The given series may be written as

$y=\sqrt{\sin{x}+y}$
Squaring both the sides, we get

$y^2=\sin{x}+y$
Differentiating both sides w.r.t.

$x$

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

$2y\displaystyle \frac { dy }{ dx } -\frac { dy }{ dx } =\cos { x }$

$\displaystyle\frac{dy}{dx}(2y-1)=\cos{x}$

$\displaystyle\frac{dy}{dx}=\frac{\cos{x}}{2y-1}$

$y=e^x \tan x + x\cdot \log_ex$ , then find

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

0%
$\displaystyle \frac {dy}{dx}=e^x \tan x+(\log x+1)$
0%
$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+x)$
0%
$\displaystyle \frac {dy}{dx}=e^x \tan x+1$
0%
$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+1)$

Explanation

$y=e^x \tan x+x\cdot \log_ex$
On differentiating, we get

$\displaystyle \frac {dy}{dx}=\frac {d}{dx}(e^x \tan x)+\frac {d}{dx}(x \log x)$

$=\tan { x } \displaystyle \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \left( \tan { x } \right) +x\frac { d }{ dx } \left( \log { x } \right) +\log { x } \frac { d }{ dx } x$

$=e^x\cdot \tan x+e^x\cdot \sec^2x+1\cdot \log x+x\cdot \displaystyle \frac {1}{x}$
Hence,

$\displaystyle \frac {dy}{dx}=e^x(\tan x+\sec^2x)+(\log x+1)$

$y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}$ , then find

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

$\displaystyle x=\frac{\pi}{4}$ .

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

Explanation

Taking

$\log$ on both sides, we get

$\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}$
Again taking

$\log$ on both the sides

$\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}$
Differentiating w.r.t.

$x$ , we get

$\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x } \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x } } \right) +\frac { d }{ dx } \left( \log { \log { \tan { x } } } \right) }$

$\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x } }{ \tan { x } } +\frac { \sec ^{ 2 }{ x } }{ \tan { x } .\log { \tan { x } } } }$

$\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x } } } } \right] }$
At

$\displaystyle x=\frac{\pi}{4}$ ,

$y=1$ and

$\log{y}=0$
So, putting this values in the equation of

$\displaystyle \frac { dy }{ dx }$ , we get

$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 3 - MCQExams.com

0:0:3

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 3 - MCQExams.com

0:0:3

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.