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

$$(\sin x)^x(\ln (\sin x)+x\cot x)$$

$$(\sin x)^x(\ln (\sin x)+x\tan x)$$

$$(\sin x)^x(\ln (\sin x)-\cot x)$$

If $$y= \log_e(x+\log_e(x+ ....)),$$ then $$\dfrac{dy}{dx}$$ at $$(x= e^2-2, y= \sqrt2)$$ is

$$\dfrac{\log2}{2\sqrt2(e^2-1)}$$

$$\dfrac{\sqrt2\log\dfrac{e}{2}}{(e^2-1)}$$

If $$ax^{2}+2hxy+by^{2}=1$$ then $$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ is equal to

$$\dfrac { { ab-h }^{ 2 } }{ \left( hx+by \right) ^{ 2 } } $$

$$\dfrac { { h }^{ 2 }-ab }{ \left( hx+by \right) ^{ 2 } } $$

$$\dfrac { { h }^{ 2 }+ab }{ \left( hx+by \right) ^{ 2 } } $$

$$\dfrac { { h }^{ 2 }-ab }{ \left( hy+by \right) ^{ 2 } } $$

If $$y=\ x^{2}\ sin\ x ,\ then\ \dfrac{dy}{dx}$$ will be

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

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 10

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

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

0%
$(\sin x)^x(\ln (\sin x)+x\cot x)$
0%
$(\ln (\sin x)+x\cot x)$
0%
$(\sin x)^x(\ln (\sin x)+x\tan x)$
0%
$(\sin x)^x(\ln (\sin x)-\cot x)$

Explanation

given

$y=(\sin x)^x$

applying

$\ln$ on both sides we get

$\ln y=x\ln(\sin x)$

differentiating both sides wrt

$x$

$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d}{dx}\left(x\ln(\sin x)\right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\ln(\sin x)+x\dfrac{1}{\sin x}\cos x$

$\dfrac{dy}{dx}=y\left(\ln(\sin x)+x\cot x\right)$

$\therefore \dfrac{dy}{dx}=(\sin x)^x\left(x\cot x\right) + (\sin x)^x \ln(\sin x)$

Solve:
If

$y=\log x^x,$ then

$\dfrac{dy}{dx}=$

0%
$1$
0%
$\log x$
0%
$\log (ex)$
0%
None of these

State True or False:

$y\, = \,x\sqrt {{a^2}\, + \,{x^2}\,} \, + \,{a^2}\,\log \left( {x\, + \,\sqrt {{a^2}\, + \,{x^2}} } \right)\,\,then\,\,\dfrac{{dy}}{{dx}}\, = \,2\sqrt {{a^2}\, + \,{x^2}}$ .

0%
True
0%
False

Explanation

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

$\dfrac{dy}{dx}=\dfrac{x}{2\sqrt{{a}^{2}+{x}^{2}}}\dfrac{d}{dx}\left({a}^{2}+{x}^{2}\right)+\sqrt{{a}^{2}+{x}^{2}}+\dfrac{{a}^{2}}{x+\sqrt{{a}^{2}+{x}^{2}}}\dfrac{d}{dx}\left(x+\sqrt{{a}^{2}+{x}^{2}}\right)$

$=\dfrac{2{x}^{2}}{2\sqrt{{a}^{2}+{x}^{2}}}+\sqrt{{a}^{2}+{x}^{2}}+\dfrac{{a}^{2}}{x+\sqrt{{a}^{2}+{x}^{2}}}\left(1+\dfrac{2x}{2\sqrt{{a}^{2}+{x}^{2}}}\right)$

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

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

$=\dfrac{{x}^{2}+{a}^{2}}{\sqrt{{a}^{2}+{x}^{2}}}\times\dfrac{\sqrt{{a}^{2}+{x}^{2}}}{\sqrt{{a}^{2}+{x}^{2}}}+\sqrt{{a}^{2}+{x}^{2}}$

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

$=\sqrt{{a}^{2}+{x}^{2}}+\sqrt{{a}^{2}+{x}^{2}}=2\sqrt{{a}^{2}+{x}^{2}}$

Hence the given statement is true.

$\dfrac{d}{dx}\displaystyle\int\limits_{f(x)}^{g(x)}h(t)dt=$

0%
$g'(x)h(g(x))$
0%
$h(g(x))-h(f(x))$
0%
$h(g(x)).g'(x)-h(f(x)).f'(x)$
0%
none of these

Explanation

We have from the Leibnitz's rule for differentiation under integral sign,

$\dfrac{d}{dx}\displaystyle\int\limits_{f(x)}^{g(x)}h(t)dt$

$=h(g(x)).g'(x)-h(f(x)).f'(x)$ .

The differential equation of all parabolas having their axis of symmetry coinciding with the axis of X is?

0%
$y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$
0%
$x\dfrac{d^2x}{dy^2}+\left(\dfrac{dx}{dy}\right)^2=0$
0%
$y\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$
0%
None of these

$y= \log_e(x+\log_e(x+ ....)),$ then

$\dfrac{dy}{dx}$ at

$(x= e^2-2, y= \sqrt2)$ is

0%
$\dfrac{1}{e^{\sqrt2}-1}$
0%
$\dfrac{\log2}{2\sqrt2(e^2-1)}$
0%
$\dfrac{\sqrt2\log\dfrac{e}{2}}{(e^2-1)}$
0%
None of these

Explanation

Given,

$\log _{e}(x+\log _{e}^{(x+…)})$

$\therefore y= \log _{e}^{x+y}$ $\because$ (as goes on to infinity)

$e^{y} = x+y$

Differentiate on both sides

$e^{y}\dfrac{dy}{dx}= 1+ \dfrac{dy}{dx}$

$\because \dfrac{dy}{dx}= \dfrac{1}{e^{y}-1}$

$\because at (x=e^{2}-2, y=\sqrt {2}) =\dfrac{1}{e^{\sqrt{2}}-1}$

Find

$\dfrac {dy}{dx}$ for

$y{e^{{x^2}}} + \tan x + \log (\sin x) = 0$

0%
None of these
0%
$- \left( {\dfrac{{y{e^{{x^2}}}.2x - {{\sec }^2}x + \cot x}}{{{e^{{x^2}}}}}} \right)$
0%
$- \left( {\dfrac{{y{e^{{x^2}}}.2x + {{\sec }^2}x + \cot x}}{{{e^{{x^2}}}}}} \right)$
0%
$\left( {\dfrac{{y{e^{{x^2}}}.2x + {{\sec }^2}x + \cot x}}{{{e^{{x^2}}}}}} \right)$

Explanation

Given

$y{e^{{x^2}}} + \tan x + \log (\sin x) = 0$

Using product rule and chain rule:

$\Rightarrow ye^{x^2}2x+\dfrac{dy}{dx}e^{x^2}+\sec^2x+\dfrac{\cos x}{\sin x}=0$

$\Rightarrow ye^{x^2}2x+\dfrac{dy}{dx}e^{x^2}+\sec^2x+\cot x=0$

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

Differentiate

$10^{x^x}+x^{x^{10}}+x^{10^x}$ w.r.t

$x$ .

0%
$10^{x^{x}}x^{x} (\log x + 1 )+ x^{10^{x}}x^{9}\left(1+10 \log x\right)+ x^{10^{x}}10^{x}\left(\cfrac{1}{x}+ \log x\right)$
0%
$(1+\log x)+x^{x^{10}}x^9(1+10\log x)+x^{10^x}10^x\left(\dfrac{1}{x}+\log x\right)$
0%
$10^{x^x}(1+\log x)+x^{x^{10}}x^9(1+10\log x)+\left(\dfrac{1}{x}+\log x\right)$
0%
$10^{x^x}(1+\log x)+(1+10\log x)+x^{10^x}10^x\left(\dfrac{1}{x}+\log x\right)$

Explanation

Let $y = 10^{x^{x}}+x^{x^{10}} +x^{10^{x}}$

Let $h=10^{x^{x}}$

Differentiate on both sides w.r.t $x$

$\cfrac{dh}{dx} =10^{x^{x}} \log (10).\cfrac{d(x^{x})}{dx}$

$\cfrac{dh}{dx} =10^{x^{x}} x^{x}(\log x + 1 )$ $(\because log 10 =1)$

Let $t = x^{x^{10}}$

Apply log on both sides

$\log t = x^{10} \log x$

Differentiate on both sides with respect to $x$

$\cfrac{1}{t}\cfrac{dt}{dx}=10x^{9} \log x + x^{9}$

$\cfrac{dt}{dx}=x^{x^{10}}x^{9}(1+ 10\log x )$

Let $m= x^{10^{x}}$

Apply log on both sides

$\log m=10^{x}\log x$

Differentiate both sides w.r.t $x$

$\cfrac{1}{m}\cfrac{dm}{dx}=(10^{x} \log 10)\log x +\cfrac{10^{x}}{x}$

$\therefore \cfrac{dm}{dx}= x^{10^{x}}(10^{x})\left(\cfrac{1}{x}+ \log x\right)$

$y= h+t+m$

Differentiating on both sides

$\cfrac{dy}{dx}=\cfrac{dh}{dx}+\cfrac{dt}{dx}+\cfrac{dm}{dx}$

$\therefore \cfrac{dy}{dx} =10^{x^{x}}x^{x} (\log x + 1 )+ x^{10^{x}}x^{9}\left(1+10 \log x\right)+ x^{10^{x}}10^{x}\left(\cfrac{1}{x}+ \log x\right)$

The shortest distance between line

$y-x=1$ and curve

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

0%
$\dfrac {\sqrt {3}}{4}$
0%
$\dfrac {3\sqrt {2}}{8}$
0%
$\dfrac {8}{2\sqrt {2}}$
0%
$\dfrac {4}{\sqrt {3}}$

Explanation

Line is

$y-x=1$ and curve

$x=y^2$

$y^2=x$

$2y\dfrac{dy}{dx}=1$

$\dfrac{dy}{dx}=\dfrac{1}{2y}=1$

$\therefore$

$y=\dfrac{1}{2}$

$x=\dfrac{1}{4}$

Tangent at

$\left(\dfrac{1}{4},\,\dfrac{1}{2}\right)$

$\dfrac{1}{2}y=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)$

$y=x+\dfrac{1}{4}$

Distance

$=\left|\dfrac{1-\dfrac{1}{4}}{\sqrt{2}}\right|=\dfrac{3}{4\sqrt{2}}=\dfrac{3\sqrt{2}}{8}$

Let

$\sqrt { x+\sqrt { x+\sqrt { x+.......\infty } } }$ then

$\dfrac { dy }{ dx } =$

0%
$\dfrac {1}{2y-1}$
0%
$\dfrac {x}{x+2y}$
0%
$\dfrac {1}{\sqrt {1+4x}}$
0%
$\dfrac {y}{2x+y}$

Explanation

Let

$y = \sqrt{x + \sqrt{x + \sqrt{x +.......\infty}}}$

$\Rightarrow \; y = \sqrt{x + \left( \sqrt{x + \sqrt{x +.......\infty}} \right)}$

$\Rightarrow \; y = \sqrt{x + y} ....... \left( 1 \right)\; \left[ \because \; y = \sqrt{x + \sqrt{x + \sqrt{x +.......\infty}}} \right]$

Squaring

${eq}^{n} \left( 1 \right)$ both sides, we have

$\Rightarrow \; {y}^{2} = x + y .......\left( 2 \right)$

Differentiating

${eq}^{n} \left( 2 \right)$ w.r.t. x, we have

$\cfrac{d}{dx} \left( {y}^{2} \right) = \cfrac{d}{dx} \left( x \right) + \cfrac{d}{dx} \left( y \right)$

$\Rightarrow \; 2y \cfrac{dy}{dx} = 1 + \cfrac{dy}{dx}$

$\Rightarrow \; 2y \cfrac{dy}{dx} - \cfrac{dy}{dx} = 1$

$\Rightarrow \; \cfrac{dy}{dx} \left( 2y - 1 \right) = 1$

$\Rightarrow \; \cfrac{dy}{dx} = \cfrac{1}{\left( 2y - 1 \right)}$

Hence, the correct answer is

$\cfrac{1}{2y - 1}$ .

Let

$y=a\cos t+b\sin t$ then

$\dfrac{d^2y}{dt^2}=$

0%
$y$
0%
$-y$
0%
$2y$
0%
none of these

Explanation

Given,

$y=a\cos t+b\sin t$ ......(1).

Now differentiating both sides with respect to

$t$ we get,

$\dfrac{dy}{dt}=-a\sin t+b\cos t$

Again differentiating both sides with respect to

$t$ we get,

$\dfrac{d^2y}{dt^2}=-a\cos t-b\sin t$

or,

$\dfrac{d^2y}{dt^2}=-y$ . [ Using (1)]

Let

$f(x)=\dfrac{1}{ax+b}$ then

$f''(0)=$

0%
$\dfrac{2a^3}{b^2}$
0%
$\dfrac{2a^2}{b^3}$
0%
$\dfrac{2a^3}{b^3}$
0%
none of these

Explanation

Given,

$f(x)=\dfrac{1}{ax+b}$

Now differentiating both sides with respect to

$x$ we get,

$f'(x)=-\dfrac{a}{(ax+b)^2}$

Again differentiating both sides with respect to

$x$ we get,

$f''(x)=\dfrac{2a^2}{(ax+b)^3}$

$f''(0)=\dfrac{2a^2}{b^3}$ .

$f(x)=\dfrac{a^x}{x^a}$ then

$f'(a)=$ ?

0%
$log a-1$
0%
$log a-a$
0%
$a log a-a$
0%
$a log a+a$

Explanation

According to question,

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

$\Rightarrow f' (a)= \dfrac{ a^{a} loga. a^{a} - a^{a+1} a^{a} }{ a^{2a} }$

Hence,

$\Rightarrow f' (a)=loga-a$

$y=2^{ax}$ and

$\dfrac{dy}{dx}=log 256$ at

$x=1$ , then the value of a is?

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

Explanation

$y=2^{ax}$

$\Rightarrow log^y_2=ax$

$log y=(log 2)ax$

$\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=alog 2$

$\Rightarrow \dfrac{1}{2^a}\times log_{256}=alog_2\Rightarrow \dfrac{1}{2^a}\times 8log_2=alog_2$

$8=a\times 2^a$

$\therefore a=2$ .

$y=\tan^{-1}(\sec x+\tan x)$ then

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

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

$y=(x^{x})^{x}$ then

$\dfrac {dy}{dx}=$

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

Explanation

We have,

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

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

On taking $\log$ both sides, we get

$\log y={{x}^{2}}\log x$ …… (1)

On differentiating w.r.t $x$ , we get

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

$\dfrac{1}{y}\dfrac{dy}{dx}=x+2x\log x$

$\dfrac{dy}{dx}=y\left( x+2x\log x \right)$

$\dfrac{dy}{dx}={{\left( {{x}^{x}} \right)}^{x}}\left( x+2x\log x \right)$

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

Hence, this is the answer.

$y=\log\left(\dfrac {1+x}{1-x}\right)^{1/4}-\dfrac {1}{2}\tan^{-1}x$ , then

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

0%
$\dfrac {x^{2}}{1-x^{4}}$
0%
$\dfrac {2x^{2}}{1-x^{4}}$
0%
$\dfrac {2x^{2}}{2(1-x^{4})}$
0%
$None\ of\ these$

${ x }^{ y }={ e }^{ x-y }$ then

$\frac { dy }{ dx } =\frac { logx }{ (1-logx)^{ 2 } }$ .

0%
True
0%
False

Explanation

${x^y} = {e^{x - y}}$

taking log to the both sides,

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

As,

$\log {a^b} = b\log a$

So,

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

$y\log x = x - y$

differentiating with respect to x,

$\begin{array}{c}\frac{{dy}}{{dx}}\log x + \frac{y}{x} = 1 - \frac{{dy}}{{dx}}\\\frac{{dy}}{{dx}}\left( {\log x + 1} \right) = 1 - \frac{y}{x}\\\frac{{dy}}{{dx}}\left( {\log x + 1} \right) = \frac{{x - y}}{x}\\\frac{{dy}}{{dx}}\left( {\log x + 1} \right) = \frac{{y\log x}}{x}\\\frac{{dy}}{{dx}} = \frac{{y\log x}}{{x\left( {\log x + 1} \right)}}\end{array}$

Thus, the equation is false

$ax^{2}+2hxy+by^{2}=1$ then

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ is equal to

0%
$\dfrac { { ab-h }^{ 2 } }{ \left( hx+by \right) ^{ 2 } }$
0%
$\dfrac { { h }^{ 2 }-ab }{ \left( hx+by \right) ^{ 2 } }$
0%
$\dfrac { { h }^{ 2 }+ab }{ \left( hx+by \right) ^{ 2 } }$
0%
$\dfrac { { h }^{ 2 }-ab }{ \left( hy+by \right) ^{ 2 } }$

Explanation

$ax^2+2hxy+6y^2=1$

$f(x, y)=ax^2+2hxy+6y^2-1=0$ .

Difference w.r.t.

$x$

$2ax+2h(xy'+y)+26yy'-1=0$

$y'=\dfrac{(ax+hy)}{hx+by}$

Difference w.r.t.

$x$

$y''=-\dfrac{(hx+6y)(a+hy')+(ax+hy)(h+6y')}{(hx+6y)^2}$

$y''=\dfrac{(ax+hy)(h+6y')-(hx+6y)(a+h'y)}{(hx+6y)^2}$

Putting

$y'=\dfrac{-(ax+hy)}{hx+6y}$

$y''$ simplified comes to be

$y''=\dfrac{ab-h^2}{(hx+6y)^2}$

$\dfrac{d^2y}{dx^2}=\dfrac{ab-h^2}{(hx+6y)^2}$ .

$f(x) = |x^2 - 5x + 6|$ , then

$f'(x)$ equals

0%
$2x - 5$ for $2 < x < 3$
0%
$5 - 2x$ for $2 < x <3$
0%
$2x - 5$ for $x > 2$
0%
$5 - 2x$ for $x < 3$

Explanation

$f(x)=|x^2-5x+6|$ ,

$f'(x)=?$

$f(x)=|(x-2)(x-3)|$

$f(1)=|1-5+6|=2$

So in

$(2, 3)$ it is negative

$\Rightarrow f(x)=(5x-x^2+6)$ in

$(2, 3)$

$f'(x)=5-2x$ for

$(2 < x < 3)$ .

$y=\ x^{2}\ sin\ x ,\ then\ \dfrac{dy}{dx}$ will be

0%
$x ^{2}\ cos\ x \ +\ 2x \ sin\ x$
0%
$2x \ sin\ x$
0%
$x ^{2}\ cos\ x$
0%
$2x \ cos\ x$

Explanation

Given

$y=x^2\sin x$

Using

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

Differentiating on both sides

$\implies \dfrac{dy}{dx}=\dfrac{d}{dx}(x^2\sin x)$

$\implies \dfrac{dy}{dx}=\sin x \dfrac{d}{dx}(x^2)+x^2\dfrac{d}{dx}(\sin x)$

$\implies \dfrac{dy}{dx}=2x\sin x+x^2\cos x$

The point(s) on the curve

${y^3} + 3{x^2} = 12y$ where the tangent is vertical, is (are)

0%
$\left( { \pm \frac{4}{{\sqrt 3 \,}},\, - 2} \right)$
0%
$\left( { \pm \frac{{\sqrt {11} }}{{3\,}},\,1} \right)$
0%
$\left( {0,\,0} \right)$
0%
$\left( { \pm \frac{4}{{\sqrt 3 \,}},\,2} \right)$

Explanation

The given equation is,

$y^3+3x^2=12y$ ------- ( 1 )

Differentiate above equation,

$3y^2\dfrac{dy}{dx}+6x=12\dfrac{dy}{dx}$

$\Rightarrow$

$(3y^2-12)\dfrac{dy}{dx}=-6x$

$\Rightarrow$

$\dfrac{dy}{dx}=\dfrac{-6x}{3y^2-12}$ ---------- ( 2 )

Let

$(x_1,\,y_1)$ be the point on the curve where the tangent is vertical to it.

Now substituting the points

$(x_1,\,y_1)$ in equation ( 2 ),

$\dfrac{dy}{dx}=\dfrac{-6x_1}{3y_1^2-12}$

Also given that the tangent at this point is verical to it.

That is the slope of the tangent is infinity at this point.

$\Rightarrow$

$3y_1^2-12=0$

$\Rightarrow$

$3y_1^2=12$

$\Rightarrow$

$y_1^2=4$

$\Rightarrow$

$y_1=\pm2$

Substituting the value of

$y_1$ in equation ( 1 ), we have,

$y_1^3+3x_1^2=12y_1$

$\Rightarrow$

$(2)^3+3x_1^2=12y_1$

$\Rightarrow$

$8+3x_1^2=12(2)$

$\Rightarrow$

$3x_1^2=24-8$

$\Rightarrow$

$3x_1^2=16$

$\Rightarrow$

$x_1^2=\dfrac{16}{3}$

$\therefore$

$x_1=\pm\dfrac{4}{\sqrt{3}}$

As the square root of a negative number is imaginary, we can neglect the value

$y_1=-2$

$\therefore$ The required point on the curve is

$\left(\pm\dfrac{4}{\sqrt{3}},\,2\right)$

$f(x)=\sqrt {(x+2\sqrt {2x-4})}+\sqrt {(x-2\sqrt {2x-4})}$ , then find the value of

$72\ f'(66)$

0%
$8$
0%
$9$
0%
$11$
0%
$22$

$x^p.y^q=(x+y)^{p+q}$ then

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

0%
$\dfrac{y}{x}$
0%
$-\dfrac{y}{x}$
0%
$\dfrac{x}{y}$
0%
$-\dfrac{x}{y}$

Explanation

Given,

${ x }^{ p }\cdot { y }^{ q }={ (x+y) }^{ p+q }$

taking

$\log$ on both sides we get,

$\log { ({ x }^{ p }\cdot { y }^{ q }) } =\log { \left[ { (x+y) }^{ p+q } \right] } \\ \Rightarrow \log { { x }^{ p } } +\log { { y }^{ q } } =(p+q)\log { (x+y) } \\ \Rightarrow p\log { x } +q\log { y } =(p+q)\log { (x+y) }$

Differentiating both sides with respect to

$x$ we get,

$p\times \dfrac { 1 }{ x } +q\times \dfrac { 1 }{ y } \dfrac { dy }{ dx } =(p+q)\dfrac { 1 }{ (x+y) } (1+\dfrac { dy }{ dx } )\\ \Rightarrow \dfrac { p }{ x } +\dfrac { q }{ y } \dfrac { dy }{ dx } =\dfrac { (p+q) }{ (x+y) } +(\dfrac { p+q }{ x+y } )\dfrac { dy }{ dx } \\ \Rightarrow \dfrac { dy }{ dx } \left[ \dfrac { p+q }{ x+y } -\dfrac { q }{ y } \right] =\dfrac { p }{ x } -(\dfrac { p+q }{ x+y } )\\ \Rightarrow \dfrac { dy }{ dx } \left[ \dfrac { py+qy-qx-qy }{ y(x+y) } \right] =\dfrac { px+py-px-qx }{ x(x+y) } \\ \Rightarrow \dfrac { dy }{ dx } \dfrac { (py-qx) }{ y(x+y) } =\dfrac { py-qx }{ x(x+y) } \\ \Rightarrow \dfrac { dy }{ dx } =\dfrac { y }{ x } \\ \therefore \dfrac { dy }{ dx } =\dfrac { y }{ x } .$

$y=\sqrt { \tan { x+\sqrt { \tan { x+\sqrt { \tan { x+.....\infty } } } } } }$ then

$\dfrac { dy }{ dx } =$

0%
$\dfrac { \cos { ^{ 2 }x +1} }{ 2y}$
0%
$\dfrac { \sec { ^{ 2 }x+1 } }{ 2y }$
0%
$\dfrac { \tan { x } +1}{ 2y }$
0%
$\dfrac { \cot { x } +1}{ 2y}$

Explanation

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

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

$y^{2}-y=\tan x$

$=02\left(\dfrac{dy}{dx}\right)y-1=\sec^{2}x$

$2\left(\dfrac{dy}{dx}\right)y=\dfrac{\sec^{2}x+1}{2}$

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

The line

$\dfrac{x}{a}+\dfrac{y}{b}=2$ tangent to

$\left ( \dfrac{x}{a} \right )^{n}\left ( \dfrac{y}{b} \right )^{n}=$ at (a,b) then

$\epsilon$ ?

0%
Z
0%
R-Z
0%
N
0%
R-{o}

Explanation

We have,

Equation of given curve is

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

Differentiation both side with respect to x and we get,

$\dfrac{n}{a}{{\left( \dfrac{x}{a} \right)}^{n-1}}+\dfrac{n}{b}{{\left( \dfrac{y}{b} \right)}^{n-1}}\dfrac{dy}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\dfrac{n}{a}{{\left( \dfrac{x}{a} \right)}^{n-\grave{\ }}}}{\dfrac{n}{b}{{\left( \dfrac{y}{b} \right)}^{n-1}}}$

$\Rightarrow {{\left( \dfrac{dy}{dx} \right)}_{\left( a,b \right)}}=\dfrac{-b}{a}$

Then, equation of tangent at point (a, b) is,

$\left( y-b \right)=\dfrac{-b}{a}\left( x-a \right)$

$\Rightarrow ay-ab=-bx+ab$

$\Rightarrow bx+ay=2ab$

$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=2$

Thus, the given curve touches the line

$\dfrac{x}{a}+\dfrac{y}{b}=2$

At point $\left( a,\,b \right)$ $\forall \,n\in N$

Hence this is the answer.

$y=|\cos x|+|\sin x|$ , then

$\dfrac {dy}{dx}$ at

$x=\dfrac {2\pi}{3}$ is

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$\dfrac {1-\sqrt {3}}{2}$
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$0$
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$\dfrac {\sqrt {3}-1}{2}$
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$\dfrac {\sqrt {3}+1}{2}$

Explanation

$y = |cos\,x|+|sin\,x|$

$, x = \dfrac{2n}{3}$

$y = -cos\,x+sin\,x$

$\left \{ x\sum (\dfrac{\pi }{2},\pi ) \right \}$

$y^{1} = sinx+cosx$

$y^{1} = sin(\dfrac{2n}{3})+cos(\dfrac{2n}{3})$

$= \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} = \dfrac{\sqrt{3}-1}{2}$ Option C is correct

1142221_1143670_ans_1376f7fff69944e79e11d99ebfe127e2.jpg

$x^{m}.y^{n}=(x+y)^{m+n}$ then

$\dfrac {dy}{dx}=$

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$\dfrac {y}{x}$
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$-\dfrac {y}{x}$
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$\dfrac {my}{x}$
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$\dfrac {ny}{x}$

Explanation

$x^my^n=(x+y)^{m+n}$

Taking log both sides

$log(x^my^n)=log(x+y)^{m+n}$

or,

$m log x+n log y=(m+n)log(x+y)$

Differentiating both sides w.r.t. x-

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

or,

$\dfrac{m}{n}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{m+n}{x+y}+\dfrac{m+n}{x+y}\dfrac{dy}{dx}$

or,

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

or,

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

or,

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

or,

$\dfrac{dy}{dx}\dfrac{1}{y}=\dfrac{1}{x}$

or,

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

$(1)$

Again differentiating both sides wrt x-

$\dfrac{d^2y}{dx^2}=\dfrac{x\dfrac{dy}{dx}-y.1}{x^2}$

or,

$\dfrac{d^2y}{dx^2}=\dfrac{x\dfrac{y}{x}-y}{x^2}$ [Using

$(1)$ ]

or,

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

or,

$\dfrac{d^2y}{dx^2}=0$ .

1184807_1154337_ans_7bc46ab2a41f4bf7a54662aefb2449a7.jpg

$x^4 + 7x^2y^2 + 9y^4 = 24 xy^3$ , then

$\dfrac{dy}{dx} =$

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$\dfrac{x}{y}$
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$\dfrac{y}{x}$
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$- \dfrac{x}{y}$
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$-\dfrac{y}{x}$

Explanation

$x^4+7x^2y^2+9y^4=24xy^3$

$4x^3+14xy^2+14x^2y\frac{dy}{dx}+36y^3\frac{dy}{dx}=24y^3+72xy^2\frac{dy}{dx}$

$\dfrac{dy}{dx}\left ( 14x^2y+36y^3-72xy^2 \right )=24y^3-4x^3-14xy^2$

$\dfrac{dy}{dx}=\dfrac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}$

$\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\dfrac{x}{y}\cdot \dfrac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}$

$\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\dfrac{24xy^3-4x^4-14x^2y^2}{14x^2y^2+36y^4-72xy^3}$

Substitute

$x^4=24xy^3-7x^2y^2-9y^4$

$\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\dfrac{24xy^3-4(24xy^3-7x^2y^2-9y^4)-14x^2y^2}{14x^2y^2+36y^4-72xy^3}$

$\dfrac{dy}{dx}=\dfrac{y}{x}\cdot\left ( \dfrac{14x^2y^2+36y^4-72xy^3}{14x^2y^2+36y^4-72xy^3} \right )$

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

$if\,x\sqrt {1 + y} + y\sqrt {1 + x} = 0,\,then\,\dfrac{{dy}}{{dx}}is\,equal\,to\,$

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$\dfrac{1}{{{{(1 + x)}^2}}}$
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$- \,\dfrac{1}{{{{(1 + x)}^2}}}$
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$\,\,\dfrac{1}{{(1 + {x^2})}}\,\,$
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$\dfrac{1}{{(1 + x)}}$

Explanation

Solution :-

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

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

squaring both sides, we get

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

$\Rightarrow x^{2}+x^{2}.y = y^{2}+y^{2}x$

$\Rightarrow x^{2}-y^{2} = y^{2}x-x^{2}y$

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

$\Rightarrow x+y = -xy$

$\Rightarrow x = -xy -y$

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

$\Rightarrow y = \dfrac{-x}{x+1}$

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

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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