If $$f(x)=\left| \log { \left| x \right| } \right| $$, then

$$f(x)$$ is continuous for all $$x$$ in its domain but not differentiable at $$x=\pm 1$$

$$f(x)$$ is continuous and differentiable for all $$x$$ in its domain

$$f(x)$$ is neither continuous nor differentiable at $$x=\pm 1$$

If $${ y }^{ x }-{ x }^{ y }=1$$, then the value of $$\cfrac { dy }{ dx } $$ at $$x=1$$ is

$$2\left( 1-\log { 2 } \right) $$

$$2\left( 1+\log { 2 } \right) $$

If $$ y^x = 2^x , $$ then $$ \dfrac {dy}{dx} $$ is equal to :

$$ \dfrac {y}{x} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {2}{y} \right) $$

$$ \dfrac {y}{x} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {x}{y} \log \left( \dfrac {y}{2} \right) $$

$$ \dfrac {y}{x} \log \left( 2y \right) $$

If $$(f(x))^{g(y)} = e^{f(x) - g(y)}$$ then $$\dfrac {dy}{dx} =$$.

$$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{2}}$$

$$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{3}}$$

$$\dfrac {f^{1}(x).\log f(x)}{g^{-1}(y) (1 - \log f(x))^{2}}$$

$$\dfrac {f^{1}(x)\log f(x)}{g(y) (1 + \log f(x))^{2}}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 9

$y = 4x - 5$ is a tangent to the curve

$y^{2} = px^{3} + q$ at

$(2, 3)$ , then

$(p + q)$ is equal to

0%
$-5$
0%
$5$
0%
$-9$
0%
$9$
0%
$0$

Explanation

Given curve is

$y^{2} = px^{3} + q .... (i)$
On differentiating w.r.t.

$x$ , we get

$2y \dfrac {dy}{dx} = 3px^{2} \Rightarrow \dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$
At

$(2, 3), \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$
Since, line

$y = 4x - 5$ is a tangent to the given curve.
Therefore,

$\text{slope} = 4 = 2p$

$\Rightarrow p = 2$
Now, put the value of

$p, x$ and

$y$ in Eq. (i), we get

$(3)^{2} = (2)(2)^{3} + q$

$\Rightarrow 9 = 16 + q$

$\Rightarrow q = -7$
Hence,

$p + q = 2 - 7 = -5$ .

$f(x)=\left| \log { \left| x \right| } \right|$ , then

0%
$f(x)$ is continuous and differentiable for all $x$ in its domain
0%
$f(x)$ is continuous for all $x$ in its domain but not differentiable at $x=\pm 1$
0%
$f(x)$ is neither continuous nor differentiable at $x=\pm 1$
0%
None of the above

Explanation

It is evident from the graph of

$f(x)=\left| \log { \left| x \right| } \right| \quad$ that

$f(x)$ is everywhere continuous but not differentiable at

$x=\pm 1$

$\sqrt{x} +\sqrt{y} = 10$ , find

$\dfrac{dy}{dx}$ at

$y = 4$ .

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

Explanation

Given equation is

$\sqrt{x}+\sqrt{y}=10$

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

$\Rightarrow \dfrac {dx}{dy}=-\dfrac {\sqrt{x}}{\sqrt{y}}=-\dfrac {10-\sqrt{y}}{\sqrt{y}}$

$\Rightarrow \dfrac {dx}{dy}=\dfrac {\sqrt{y}-10}{\sqrt{y}}$

$\Rightarrow \left ( \dfrac {dx}{dy} \right )_{y=4}=\dfrac {2-10}{2}=\dfrac {-8}{2}=-4$

$xe^{xy} + ye^{-xy} = \sin^{2}x$ , then

$\dfrac {dy}{dx}$ at

$x = 0$ is

0%
$2y^{2} - 1$
0%
$2y$
0%
$y^{2} - y$
0%
$y^{2} + 1$
0%
$y^{2} - 1$

Explanation

Given,

$xe^{xy} + ye^{-xy} = \sin^{2}x$
On differentiating w.r.t.

$x$ , we get

$e^{xy} + xe^{xy} \left \{x \dfrac {dy}{dx} + y\right \} + \dfrac {dy}{dx} e^{-xy}$

$-ye^{-xy} \left \{x \dfrac {dy}{dx} + y\right \} = 2\sin x \cdot \cos x$

$\Rightarrow \left \{x^{2} e^{xy} + e^{-xy} - xe^{y} e^{-xy}\right \}\dfrac {dy}{dx} + \left \{e^{xy} + xye^{xy} - y^{2} e^{-xy}\right \} =\sin 2x$
On putting

$x = 0$ , we get

$\left \{0 + e^{-0} - 0\right \} \left (\dfrac {dy}{dx}\right )_{(x = 0)} + \left \{e^{0} + 0 - y^{2}e^{-0}\right \} = \sin 0$

$\Rightarrow \left [(1)\dfrac {dy}{dx}\right ]_{x = 0} + (1 - y^{2}) = 0$

$\Rightarrow \left [\dfrac {dy}{dx}\right ]_{x = 0} = y^{2} - 1$ .

$x^2 + 2xy + 2y^2 = 1,$ then

$\dfrac {dy}{dx}$ at the point where

$y=1$ is equal to :

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

Explanation

Given,

$x^2 + 2xy + 2y^2 = 1$
Put

$y = 1,$

$x^2 + 2x(1) + 2(1)^2 = 1$

$\Rightarrow x^2 + 2x + 1 = 0$

$\Rightarrow (x+1)^2 = 0$

$\Rightarrow x = -1$
On differentiating Eq. (i) w.r.t.

$x,$ we get

$2x + 2x \dfrac {dy}{dx} + 2y + 4y \dfrac {dy}{dx} = -0$

$\Rightarrow 2 \dfrac {dy}{dx} (x+2y) = -2 (y+x)$

$\Rightarrow \dfrac {dy}{dx} = \dfrac {-(y+x)}{x+2y}$
When

$x=-1, y = 1$

$\dfrac {dy}{dx} = - \dfrac {(1-1) }{-1+2} = 0$

${ y }^{ x }-{ x }^{ y }=1$ , then the value of

$\cfrac { dy }{ dx }$ at

$x=1$ is

0%
$2\left( 1-\log { 2 } \right)$
0%
$2\left( 1+\log { 2 } \right)$
0%
$2-\log { 2 }$
0%
$2+\log { 2 }$

Explanation

We have

$\quad { y }^{ x }-{ x }^{ y }=1....(i)$

$\Rightarrow { e }^{ x\log { y } }-{ e }^{ y\log { x } }=1$
On differentiating w.r.t.

$x$ we get

${ y }^{ x }\left( \cfrac { x }{ y } .\cfrac { dy }{ dx } +\log { y } \right) -{ x }^{ y }\left( \cfrac { dy }{ dx } \log { x } +\cfrac { y }{ x } \right) =0$
On putting

$x=1$ and

$y=2$ we get

$2\left( \cfrac { 1 }{ 2 } .\cfrac { dy }{ dx } +\log { 2 } \right) -(0+2)=0$

$\cfrac { dy }{ dx } =2-2\log { 2 }$

$y^x = 2^x ,$ then

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

0%
$\dfrac {y}{x} \log \left( \dfrac {2}{y} \right)$
0%
$\dfrac {x}{y} \log \left( \dfrac {2}{y} \right)$
0%
$\dfrac {y}{x} \log \left( \dfrac {y}{2} \right)$
0%
$\dfrac {x}{y} \log \left( \dfrac {y}{2} \right)$
0%
$\dfrac {y}{x} \log \left( 2y \right)$

Explanation

Given,

$y^x = 2^x$
On taking

$\log$ on both sides, we get

$x \log y = x \log 2$
On differentiating w.r.t.

$x,$ we get

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

$\Rightarrow \dfrac {dy}{dx} = \dfrac {y}{x} [ \log 2 - \log y ]$

$\Rightarrow \dfrac {dy}{dx}= \dfrac {y}{x} \left[ \log \dfrac{2}{y} \right]$

$y = x - x^{2}$ , then the derivative of

$y^{2} w.r.t. x^{2}$ is

0%
$2x^{2} + 3x - 1$
0%
$2x^{2} - 3x + 1$
0%
$2x^{2} + 3x + 1$
0%
None of these

Explanation

Given,

$y = x - x^{2}$
On differentiating w.r.t.

$x$ , we get

$\dfrac {dy}{dx} = 1 - 2x$
Now,

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

$= \dfrac {2y\dfrac {dy}{dx}}{2x} = 2y \dfrac {(1 -2x)}{2x}$

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

$= (1 - x)(1 - 2x)$

$= 1 - 3x + 2x^{2}$

$= 2x^{2} - 3x + 1$ .

$(f(x))^{g(y)} = e^{f(x) - g(y)}$ then

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

0%
$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{2}}$
0%
$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{3}}$
0%
$\dfrac {f^{1}(x).\log f(x)}{g^{-1}(y) (1 - \log f(x))^{2}}$
0%
$\dfrac {f^{1}(x)\log f(x)}{g(y) (1 + \log f(x))^{2}}$

Explanation

${ f\left( x \right) }^{ g\left( y \right) }={ e }^{ f\left( x \right)-g\left( y \right) }$

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

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

${ f\left( x \right) } = { g\left( y \right) }\left[ 1+\log { { f\left( x \right) } }\right]$ ...(1)

$f^{ 1 }\left( x \right) ={ g\left( y \right) }\left[ \frac { f^{ 1 }\left( x \right) }{ f\left( x \right) } \right] +g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx }$

Rearranging,

$f^{ 1 }\left( x \right) \left[ \frac { f\left( x \right) -g\left( y \right) }{ f\left( x \right) } \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx }$

from eqn. (1)

$f^{ 1 }\left( x \right) \left[ \frac { g\left( y \right) \times \log { f\left( x \right) } }{ f\left( x \right) } \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx }$

again, from eqn (1),

$f^{ 1 }\left( x \right) \left[ \frac { \log { f\left( x \right) } }{ \left[ 1+\log { { f\left( x \right) } } \right]} \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx }$

$\frac { dy }{ dx } =\frac { f^{ 1 }\left( x \right) }{ g^{ 1 }\left( y \right) } \times \left[ \frac { \log { f\left( x \right) } }{ { \left[ 1+\log { { f\left( x \right) } } \right] }^{ 2 } } \right]$

Let

$f$ and

$g$ be differential functions satisfying

$g' (a) =2 , g (a) b$ and

$fog = I$ (identify function) then

$f'(b) =$ :

0%
$1/2$
0%
$2$
0%
$2/3$
0%
None of these

Explanation

$fog = I \Rightarrow fog (x) = x \forall x$

$\Rightarrow f' (g(x)) g'(x) = 1 \forall x$

$\Rightarrow f'(g(a)) = \dfrac {1}{g'(a)} = \dfrac {1}{2} \Rightarrow f'(b) = \dfrac {1}{2}$

$\dfrac {d}{dx}\left (\sqrt {3} \sin \left (2x + \dfrac {\pi}{3}\right ) + \cos \left (2x + \dfrac {\pi}{3}\right )\right ) =$ _________.

0%
$4\cos 2x$
0%
$-4\sin 2x$
0%
$4\sin 2x$
0%
$-4\cos 2x$

Explanation

$\dfrac {d}{dx}\left (\sqrt {3} \sin \left (2x + \dfrac {\pi}{3}\right ) + \cos \left (2x + \dfrac {\pi}{3}\right )\right )$

$=\dfrac {d}{dx}\left (2\sin \left (2x + \dfrac {\pi}{3} + \dfrac {\pi}{6}\right )\right )$

$=\dfrac {d}{dx} \left (2\sin \left (2x + \dfrac {\pi}{2}\right )\right )$

$=\dfrac {d}{dx} (2\cos 2x)$

$=-4\sin 2x$ .

$|x| < 1$ , then

$\dfrac{d}{dx}\left[1+\dfrac{p}{q}x+\dfrac{p(p+q)}{2!}\left(\dfrac{x}{q}\right)^2+\dfrac{p(p+q)(p+2q)}{3!}\left(\dfrac{x}{q}\right)^3....\infty\right]=$

0%
$\dfrac{p}{q(1-x)^{\frac{p}{q}+1}}$
0%
$\dfrac{p}{q(1-x)^{\frac{p}{q}}}$
0%
$(1-x)^{-pq-1}$
0%
$(1-x)^{pq+1}$

Explanation

$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}+\dfrac{p(p+q)}{2!}\dfrac{2x}{q^2}+\dfrac{p(p+q)(p+2q)}{3!}\dfrac{3x^2}{q^3}+\dots$

$=\dfrac{p}{q}\left(1+\dfrac{(p+q)}{1!}\dfrac{x}{q}+\dfrac{(p+q)(p+2q)}{2!}\dfrac{x^2}{q^2}+\dots\right)$

Now, consider

$(1-a)^n=1-na+\dfrac{n(n-1)}{2!}x^2+\dots$

Put

$a=x$ and

$n=-\left(1+\dfrac{p}{q}\right)$

$\therefore (1-x)^{-\left(1+\frac{p}{q}\right)}=1+\dfrac{(p+q)}{q}x+\left[-\dfrac{(p+q)}{q}\left(-\dfrac{(p+q)}{q}-1\right)\right]\dfrac{x^2}{2!}\dots$

$=1+\dfrac{(p+q)}{q}x+\left[\dfrac{(p+q)}{q}\left(\dfrac{(p+q+q)}{q}\right)\right]\dfrac{x^2}{2!}\dots$

$=1+\dfrac{(p+q)}{q}x+\dfrac{(p+q)(p+2q)}{q^2}\dfrac{x^2}{2!}\dots$

Hence,

$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}(1-x)^{-\left(1+\frac{p}{q}\right)}$

This is the required answer.

$y = Tan^{-1} \left (\dfrac {\log (e/x^{2})}{\log ex^{2}}\right ) + Tan^{-1} \left (\dfrac {3 + 2\log x}{1 - 6\log x}\right )$ then

$\left (\dfrac {dy}{dx}\right )_{x = 2} + \left (\dfrac {dy}{dx}\right )_{x = 3}$ .

0%
$-6$
0%
$2$
0%
$0$
0%
$-2$

Explanation

$y=\tan ^{ -1 }{ \left( \cfrac { \log { \cfrac { e }{ { x }^{ 2 } } } }{ \log { e{ x }^{ 2 } } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 3+2\log { x } }{ 1-6\log { x } } \right) } \\ \quad =\tan ^{ -1 }{ \left( \cfrac { \log { e } -\log { { x }^{ 2 } } }{ \log { e } +\log { { x }^{ 2 } } } \right) } +\tan ^{ -1 }{ \left[ \cfrac { 3+2\log { x } }{ 1-3-2\log { x } } \right] } \\ \quad =\tan ^{ -1 }{ \left[ \cfrac { 1-\log { { x }^{ 2 } } }{ 1+\log { { x }^{ 2 } } } \right] + } \tan ^{ -1 }{ 3 } +\tan ^{ -1 }{ 2\log { x } } \\ \quad =\tan ^{ -1 }{ 1 } -\tan ^{ -1 }{ \log { { \left( x \right) }^{ 2 } } +\tan ^{ -1 }{ 3 } } +\tan ^{ -1 }{ \log { { \left( x \right) }^{ 2 } } } \\ \quad =\tan ^{ -1 }{ 1+\tan ^{ -1 }{ 3 } } \\ y=\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 3 }$ is constant

So, $\cfrac { dy }{ dx } =0$

Answer C

Let

$\int _{ 0 }^{ x }{ \left( \cfrac { bt\cos { 4t } -a\sin { 4t } }{ { t }^{ 2 } } \right) } dt=\cfrac { a\sin { 4x } }{ x }$ then

$a$ and

$b$ are given by

0%
$a=\cfrac { 1 }{ 4 } ,b=1$
0%
$a=2,b=2$
0%
$a=-1,b=4$
0%
$a=2,b=4$

Explanation

$\displaystyle \int_{0}^{x}{\dfrac{bt\cos 4t-a\sin 4t}{t^{2}}}dt=\dfrac{a\sin 4x}{x}$

$\text{Differentiating w.r.t. x}$

$\dfrac{bx\cos 4x-a\sin 4x}{x^{2}}=\dfrac{4ax\cos 4x-a\sin 4x}{x^{2}}$

$b=4a \Rightarrow a=\dfrac{1}{4},b=1$

$8f(x) + 6f\left( {{1 \over x}} \right) = x + 5$ and

$y = {x^2}f(x)$ ,then

${{dy} \over {dx}}$ at x=-1 is equal to

0%
0
0%
${1 \over {14}}$
0%
$- {1 \over {14}}$
0%
None of these

Explanation

$8{f(x)}+6{f\bigg(\dfrac{1}{x}\bigg)}=x+5\implies f(-1)=\dfrac{2}{7}$

Differentiating on both sides

$8{f'(x)}-\dfrac{6}{x^{2}}f'\bigg(\dfrac{1}{x}\bigg)=1$

Put

$x=-1\implies f'(-1)=\dfrac{1}{2}$

$y=x^{2}f(x)$

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

$x=-1,\dfrac{d{y}}{d{x}}=2(-1)\times \dfrac{2}{7}+(-1)^{2}\dfrac{1}{2}=-\dfrac{1}{14}$

$f:\mathrm{R}\to\mathrm{R}$ is a differentiable function such that

$f'(x)\gt 2f(x)\forall x\in \mathrm{R}$ and

$f(0)=1$ , then

0%
$f(x)$ is increasing in $(0,\infty)$
0%
$f(x)$ is decreasing in $(0,\infty)$
0%
$f(x)\gt e^{2x}$ in $(0,\infty)$
0%
$f(x)\lt e^{2x}$ in $(0,\infty)$

Explanation

Given that,

$f'\left( x \right) >2f\left( x \right) ,f\left( 0 \right) =1$

$\int { \cfrac { df\left( x \right) }{ f\left( x \right) } } >\int { 2dx }$

$\ln { f\left( x \right) } >2x+c$

$f\left( x \right) >k{ e }^{ 2x },\quad k>0$

$f\left( 0 \right) =1$

$1>k\quad \therefore 0<k<1$

$\therefore f\left( x \right) >k{ e }^{ 2x }>0$

$f'\left( x \right) >2f\left( x \right) >0$

$\therefore f\left( x \right)$ is increasing function (since

$f'\left( x \right) >0$ ) in

$\left( 0,\infty \right)$

If y=

$A\sin (\omega t - kx)$ , then the value of

${{dy} \over {dx}}$ is

0%
$A\cos (\omega t - kx)$
0%
$- A\omega \cos (\omega t - kx)$
0%
$Ak\cos (\omega t - kx)$
0%
$- Ak\cos (\omega t - kx)$

Explanation

$y=A\sin (wt-kx)$

$\therefore \dfrac{dy}{dx}=A\cos (wt-kx)x\times-k$

$\left[\because \dfrac{d}{dx}\sin x=\cos x\right]$

$=-Ak\cos (wt-kx)$

$\frac{d^n}{dx^n}[log(ax+b)]$ is equal to:

0%
$\frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$
0%
$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^{n+1}}$
0%
$\frac{(-1)^{n+1} ({n+1)}! a^{n-1}}{(ax+b)^{n+1}}$
0%
$\frac{(-1)^{n-1} {(n-1)}! a^n}{(ax+b)^n}$

Explanation

Let

$y=\log { \left( ax+b \right) }$ , let

$D=\cfrac { d }{ dx }$

$\cfrac { dy }{ dx } =Dy=\cfrac { a }{ ax+b }$

${ D }^{ 2 }y=\cfrac { -{ a }^{ 2 } }{ { \left( ax+b \right) }^{ 2 } } ,{ D }^{ 3 }y=\cfrac { 2{ a }^{ 3 } }{ { \left( ax+b \right) }^{ 3 } }$

${ D }^{ 4 }y=\cfrac { -3.2.1.{ a }^{ 4 } }{ { \left( ax+b \right) }^{ 4 } }$

$\therefore { D }^{ n }y=\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } }$

$\therefore \cfrac { { d }^{ n } }{ d{ x }^{ n } } \left( \log { \left( ax+b \right) } \right) =\cfrac { { \left( -1 \right) }^{ n-1 }\left( n-1 \right) !{ a }^{ n } }{ { \left( ax+b \right) }^{ n } }$

Derivative of

$log_e(log_e|\sin x|)$ with respect to

$x$ at

$x=\dfrac{\pi}{6}$ is

0%
$-\dfrac{\sqrt{3}}{log_e2}$
0%
$\dfrac{\sqrt{3}}{log_e2}$
0%
$-\dfrac{\sqrt{3}}{2\log2}$
0%
does not exist

Statement I: The function f(x) in the figure is differentiable at x = a
Statement II: The function f(x) continuous at x = a

0%
Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I.
0%
Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.
0%
Statement I is true but Statement II is false
0%
Statement I is false but Statement II is true.

Explanation

$f(x)$ is continuous at as it can be observed from the figure there is no discontinuous at

$x=a$ but coming to differentiability

$f(x)$ is not differentiable at

$x=a$ since the curve is sharp[i.e the left slope and right slope are not equal]

therefore the answer is statement I is false but statement II is true

$y=x^x$ then

$\dfrac{d^2y}{dx^2}-\dfrac{1}{y}\left(\dfrac{dy}{dx}\right)^2-\dfrac{y}{x}=0$

0%
True
0%
False

$e^y(x + 1) = 1$ , then

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

0%
$y$
0%
$\dfrac{dy}{dx}$
0%
$-y$
0%
$\left(\dfrac{dy}{dx}\right)^2$

Explanation

Given

$e^y(x+1)=1$

or,

$e^y=\dfrac{1}{(x+1)}$

or,

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

Now

$\dfrac{dy}{dx}=-\dfrac{1}{(1+x)}$

And

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

Given

$y=\dfrac {3}{x}, \dfrac {dy}{dx}=$

0%
$3$
0%
$\dfrac {3}{x^{2}}$
0%
$\dfrac {-3}{x^{2}}$
0%
$3x$

Explanation

$y=\dfrac{3}{x}$

differentiating with respect to x

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

State True or False.

If $\dfrac{{{e^y}}}{{{e^x}}} = xy$ , then $y' = \dfrac{{2 - \log x}}{{{{\left( {1 - \log x} \right)}^2}}}$

0%
True
0%
False

Explanation

$\dfrac{{e}^{y}}{{e}^{x}}=xy\\$

$\Rightarrow\,{e}^{y-x}=xy\\$

$\Rightarrow\,(y-x)\log{e}=\log{xy}\\$

$\Rightarrow\,y-x=\log{x}+\log{y}\\$

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

$\Rightarrow\,\dfrac{dy}{dx}\left(1-\dfrac{1}{y}\right)=1+\dfrac{1}{x}\\$

$\Rightarrow\,\dfrac{dy}{dx}\left(\dfrac{y-1}{y}\right)=\dfrac{x+1}{x}\\$

$\Rightarrow\,\dfrac{dy}{dx}=\dfrac{y\left(x+1\right)}{x\left(y-1\right)}\\$

Given statement is false.

$f(x) = \bigg[ \frac {a+x}{1+x} \bigg]^{a+1+2x}$ then

${a^{a+1}} \bigg [ 2 \ log \ a + {\frac {1-a ^2}{a}} \bigg]$ is

0%
$f^\prime (1)$
0%
$f^\prime (0)$
0%
$f^\prime (2)$
0%
$f^{\prime \prime}$

Find

$\dfrac{dy}{dx}$ of the function given:

$xy = e^{(x-y)}$

0%
$\dfrac{e^{x-y}+y}{x-e^{x-y}}$
0%
$\dfrac{e^{x-y}+x}{e^{x-y}-y}$
0%
$\dfrac{e^{x-y}-y}{x+e^{x-y}}$
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$\dfrac{e^{x-y}-x}{e^{x-y}+y}$

Explanation

Differentiating with respect to x

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

$\dfrac{d}{{dx}}\left( {{{\sec }^2}x+{{{\mathop{\rm cosec}\nolimits} }^2}x} \right) =$

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$- 4\sec x \cdot \tan x \cdot \cos ec\,x \cdot \cot x$
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$4\sec x \cdot \,\cos ec\,x$
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$2\sec \,x \cdot \,\tan x - 2\cos ec\,x \cdot \cot \,x$
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$2{\sec ^2} \cdot \tan x - 2{{\mathop{\rm cosec}\nolimits} ^2}x \cdot \cot x$

Explanation

Let,

$f(x)=\sec^2+ \cos ec^2 x$

$\dfrac{d}{dx}f(x)=2\sec x.\sec x\tan x\cos ex^2x+\sec^2x2\cos ex x(-cosec\ x\cot x)$

$=2{\sec ^2x} \cdot \tan x - 2{{\mathop{\rm cosec}\nolimits} ^2}x \cdot \cot x$

Function f(x)=

$\left| {x - 2} \right| - 2\left| {x - 4} \right|\,is$ discontinous at:

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$x=2,4$
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$x=2$
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No where
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Except $x=2$

Explanation

$f(x)=\mid x-2 \mid -2\mid x-4 \mid$

$\therefore f\left(x \right)=\begin{Bmatrix} -(x-2)+2(x-4),\qquad x<2\\ (x-2)+2(x-4), \qquad 2\le x \le 4\\ (x-2)-2(x-4) \qquad x\gt 4 \\ \end{Bmatrix}$

$x=2$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 2^{-}}=-(2-2)+2(2-4) = -4$

R.H.LRight Hand Limit/0

$\displaystyle\lim_{x\to\ 2^{+}}=(2-2)+2(2-4)=-4$

$\therefore \displaystyle\lim_{x\to\ 2^{-}}f(x) = \displaystyle\lim_{x\to\ 2^{+}}f(x)$

$x=4$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 4^{-}}=(4-2)+2(4-4)=2$

R.H.L(Right Hand Limit)

$\displaystyle\lim_{x\to\ 4^{+}}=(4-2)-2(4-4)=2$

$\therefore \displaystyle\lim_{x\to\ 4^{-}}f(x) = \displaystyle\lim_{x\to\ 4^{+}}f(x)$

$\therefore \ f(x)$ is continuous at everywhere.

Differentiate with respect to

$x$ .

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

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$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$
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$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$
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$x^{\cos x}\left[ {\cos x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec x.\log \sin x]$
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None

Explanation

let

$y=x^{\cos x} +\sin x ^{\tan x}$

let

$h=x^{\cos x}$

$\dfrac{d}{dx}(f(x)g(x))=f'(x)g(x)+f'(x)f(x)$

applying

$log$ on both sides we get

$\log h=\cos x\log x$

differentiate on both sides wrt

$x$

$\dfrac{1}{h}\dfrac{dh}{dx}=-\sin x\log x+\dfrac{\cos x}{x}$

$\therefore \dfrac{dh}{dx}=x^{\cos x}\left(\dfrac{\cos x}{x}-\sin x \log x\right)$

let

$t=\sin x^{\tan x}$

applying

$\log$ on both sides

$\dfrac{1}{t}\dfrac{dt}{dx}=\sec ^2x\log {(\sin x)}+\dfrac{\tan x}{\sin x}(\cos x)$

$\therefore \dfrac{dt}{dx}=\sin x^{\tan x}\left(1+\sec ^2x \log (\sin x)\right)$

$y=h+t$

$\implies \dfrac{dy}{dx}=\dfrac{dh}{dx}+\dfrac{dt}{dx}$

$\therefore \dfrac{dy}{dx}=x^{\cos x}\left(\dfrac{\cos x}{x}-\sin x\log x\right)+\sin x^{\tan x}\left(1+\sec ^2x\log (\sin x)\right)$

Differentiate

${x^{\tan x}} + {{\mathop{\rm tanx}\nolimits} ^x}$ with respect to

$x$

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$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
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$x^{\tan x}(\sec^2x \log \sec x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
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$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})-\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
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$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x-\dfrac{2}{\sin 2x})$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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