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 12 Commerce Maths Continuity And Differentiability Quiz 5

Let

$y=\tan ^{ -1 }{ \left( \sec { x } +\tan { x } \right) }$ . Then,

$\cfrac { dy }{ dx } =$

0%
$\cfrac { 1 }{ 4 }$
0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 1 }{ \sec { x } +\tan { x } }$
0%
$\cfrac { 1 }{ \sec ^{ 2 }{ x } }$
0%
$\cfrac { 1 }{ \tan { x } }$

Explanation

Given,

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

$\Rightarrow y=\tan ^{ -1 }{ \left[ \tan { \left( \cfrac { \pi }{ 4 } +\cfrac { x }{ 2 } \right) } \right] }$

$\left[ \because \tan { \left( \cfrac { \pi }{ 4 } +\cfrac { x }{ 2 } \right) } =\sec { x } +\tan { x } \right]$

$\Rightarrow y=\cfrac { \pi }{ 4 } +\cfrac { x }{ 2 }$

On differentiating w.r.t

$x$ we get

$\cfrac { dy }{ dx } =0+\cfrac { 1 }{ 2 } =\cfrac { 1 }{ 2 }$

$x = \sin t$ and

$y = \tan t$ , then

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

0%
$\cos^3 t$
0%
$\dfrac{1}{\cos^3 t}$
0%
$\dfrac{1}{\cos^2 t}$
0%
$\sin^2 t$
0%
$\dfrac{1}{\sin^2 t}$

Explanation

Given,

$x = \sin t$ and

$y = \tan t$
On differentiating both sides w.r.t.

$t$ respectively, we get

$\dfrac{dx}{dt} = \cos t$ and

$\dfrac{dy}{dt} = \sec^2 t$
Therefore,

$\dfrac{dy/dt}{dx/dt} = \dfrac{\sec^2 t}{\cos t} = \dfrac{1}{\cos^3 t}$

$f(x)=\cos ^{ -1 }{ \left\{ \cfrac { 1-{ \left( \log _{ e }{ x } \right) }^{ 2 } }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} }$ , then

$f'(e)$

0%
Does not exist
0%
Is equal to $\cfrac { 2 }{ e }$
0%
Is equal to $\cfrac { 1 }{ e }$
0%
Is equal to $1$

Explanation

We have

$\quad f(x)=\cos ^{ -1 }{ \left\{ \cfrac { 1-{ \left( \log _{ e }{ x } \right) }^{ 2 } }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} }$

$\quad =2\tan ^{ -1 }{ \left( \log _{ e }{ x } \right) } \quad \left[ \because \log _{ e }{ x } >0 \right]$ (in the

$nbd$ of

$x=e$ )

$\Rightarrow f'(x)=\cfrac { 2 }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } } .\cfrac { 1 }{ x }$

$\Rightarrow f'(e)=\cfrac { 2 }{ 1+{ \left( \log _{ e }{ e } \right) }^{ 2 } } .\cfrac { 1 }{ e }$

$=\cfrac { 2 }{ 1+1 } .\cfrac { 1 }{ e }$

$=\cfrac { 1 }{ e }$

$y=\tan ^{ -1 }{ \left( \cfrac { a\cos { x } -b\sin { x } }{ b\cos { x } +a\sin { x } } \right) }$ , then

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

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

Explanation

$y=\tan ^{ -1 }{ \left( \cfrac { a\cos { x } -b\sin { x } }{ b\cos { x } +a\sin { x } } \right) }$
Let

$a=r\sin { \theta } ;b=r\cos { \theta }$

$\therefore y=\tan ^{ -1 }{ \left\{ \cfrac { r\sin { \left( \theta -x \right) } }{ r\cos { \left( \theta -x \right) } } \right\} }$

$\Rightarrow y=\theta -x,y=\tan ^{ -1 }{ \left( \cfrac { a }{ b } \right) } -x$

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

$x=\cos { \theta } ,y=\sin { 5\theta }$ , then

$(1-{ x }^{ 2 })\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\cfrac { dy }{ dx } =$

0%
$-5y$
0%
$5y$
0%
$25y$
0%
$-25y$

Explanation

We have

$x=\cos { \theta } ,y=\sin { \theta }$

$\quad \therefore \cfrac { dy }{ dx } =\cfrac { \cfrac { dy }{ d\theta } }{ \cfrac { dx }{ d\theta } } =-\cfrac { 5\cos { 5\theta } }{ \sin { \theta } }$

$\Rightarrow \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =-5\cfrac { d }{ d\theta } \left( \cfrac { 5\cos { 5\theta } }{ \sin { \theta } } \right) .\cfrac { d\theta }{ dx } \quad$

$=\cfrac { -25\sin { \theta } \sin { 5\theta } -5\cos { \theta } \cos { 5\theta } }{ \sin ^{ 3 }{ \theta } }$

$\quad \therefore \left( 1-{ x }^{ 2 } \right) \cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\cfrac { dy }{ dx }$

$=\left( 1-\cos ^{ 2 }{ \theta } \right) \left[ \cfrac { -25\sin { \theta } \sin { 5\theta } -5\cos { \theta } \cos { 5\theta } }{ \sin ^{ 3 }{ \theta } } \right] -\cos { \theta } \left[ \cfrac { -5\cos { 5\theta } }{ \sin { \theta } } \right]$

$=-25\sin { 5\theta } =-25y$

$y=x\log { \left( \cfrac { x }{ a+bx } \right) }$ , then

${ x }^{ 3 }\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ is equal to

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

Explanation

From the given relation

$y=x\log { \left( \cfrac { x }{ a+bx } \right) }$

$\cfrac { y }{ x } =\log { x } -\log { (a+bx) }$
On differentiating w.r.t

$x$ , we get

$\cfrac { \left( x\cfrac { dy }{ dx } -y \right) }{ { x }^{ 2 } } =\cfrac { 1 }{ x } -\cfrac { 1 }{ a+bx } b$
Therefore,

$x\cfrac { dy }{ dx } -y=\cfrac { ax }{ a+bx }$ ...(i)
Again differentiating both sides w.r.t

$x$ we get

$x\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +\cfrac { dy }{ dx } -\cfrac { dy }{ dx } =\cfrac { (a+bx)a-axb }{ { (a+bx) }^{ 2 } }$

$\Rightarrow { x }^{ 3 }\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\cfrac { { a }^{ 2 }{ x }^{ 2 } }{ { (a+bx) }^{ 2 } } ={ \left( x\cfrac { dy }{ dx } -y \right) }^{ 2 }$ ..... (from Eq(i))

Let

$f\left( x \right)$ be a non-negative continuous function such that the area bounded by the curve

$y=f\left( x \right)$ ,

$x$ -axis and the ordinates

$x=\dfrac { \pi }{ 4 }$ and

$x=\beta > \dfrac { \pi }{ 4 }$ is

$\left( \beta \sin { \beta } +\dfrac { \pi }{ 4 } \cos { \beta } +\sqrt { 2 } \beta \right)$ . Then,

$f\left( \dfrac { \pi }{ 2 } \right)$ is

0%
$\left( 1-\dfrac { \pi }{ 4 } +\sqrt { 2 } \right)$
0%
$\left( 1-\dfrac { \pi }{ 4 } -\sqrt { 2 } \right)$
0%
$\left( \dfrac { \pi }{ 4 } -\sqrt { 2 } +1 \right)$
0%
$\left( \dfrac { \pi }{ 4 } +\sqrt { 2 } -1 \right)$

Explanation

Given,

$\displaystyle\int _{ { \pi }/{ 4 } }^{ \beta }{ f\left( x \right) dx } =\beta \sin { \beta } +\dfrac { \pi }{ 4 } \cos { \beta } +\sqrt { 2 } \beta$
On differentiating with respect to

$\beta$ on both sides, we get

$f\left( \beta \right) =\sin { \beta } +\beta \cos { \beta } -\dfrac { \pi }{ 4 } \sin { \beta } +\sqrt { 2 }$ (by Leibnitz rule)
Put

$\beta =\dfrac { \pi }{ 2 }$
Then,

$f\left( \dfrac { \pi }{ 2 } \right) =\sin { \dfrac { \pi }{ 2 } } +\dfrac { \pi }{ 2 } \cos { \dfrac { \pi }{ 2 } } -\dfrac { \pi }{ 4 } \sin { \dfrac { \pi }{ 2 } } +\sqrt { 2 }$

$= 1 + 0 - \dfrac { \pi }{ 4 } +\sqrt { 2 }$

$= 1 - \dfrac { \pi }{ 4 } +\sqrt { 2 }$

Let

$f(x)=2\tan^{-1}x+\sin^{-1}\left(\displaystyle\frac{2x}{1+x^2}\right)$ . Then

0%
$f'(2)=f'(3)$
0%
$f'(2)=0$
0%
$f'(1/2)=16/5$
0%
All of these

Explanation

We know that,

$\sin^{-1}\left(\displaystyle\frac{2x}{1+x^2}\right)=\left\{\begin{matrix} 2\tan^{-1}x, & if -1\leq x \leq 1\\ \pi -2\tan^{-1}x, & if x> 1\\ -\pi -2\tan^{-1}x, & if x < -1\end{matrix}\right.$
Therefore,

$f(x)=\left\{\begin{matrix} 4\tan^{-1} x, & if -1 \leq x \leq 1\\ \pi & if x>1 \\ -\pi, & if x < -1\end{matrix}\right.$

$\Rightarrow f'(x)=\left\{\begin{matrix} \displaystyle\frac{4}{1+x^2}, & if -1 < x <1 \\ 0, & if |x|>1\end{matrix}\right.$

$\Rightarrow f'(2)=f'(3)=0$ and

$f'\left(\displaystyle\frac{1}{2}\right)=\frac{4}{1+\displaystyle \left(\displaystyle \frac{1}{2}\right)^2}=\displaystyle \frac{16}{5}$

The value of a for which the function

$f(x)=\left\{\begin{matrix} \tan^{-1} a-3x^2, & 0<x<1 \\ -6x, & x\geq 1\end{matrix}\right.$ has a maximum at

$x=1$ , is?

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

Explanation

We have,

$\displaystyle f(x)=\left\{\begin {matrix} \tan^{-1} a-3x^2, & 0 < x < 1\\ -6x , & x\geq 1\end{matrix}\right.$
If

$f(x)$ attains a maximum at

$x=1$ , then

$f'(1)$ must exist and should be zero. This means that

$f(x)$ must be continuous and differentiable at

$x=1$ .
We observe that

$f(x)$ will be continuous at

$x=1$ , if

$\tan^{-1}a=-3$ .
But, (LHD at

$x=1$ )

$=$ (RHD at

$x=1$ )

$=-6\neq 0$ for any value of

$a$ .
Hence, there is no value of

$a$ for which

$f(x)$ has a minimum at

$x=1$ .

$y^x = 2^x ,$ 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,

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

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

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

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

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

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

Find the derivative of

$\sin(2\sin^{-1}x)$ .

0%
$\dfrac{2\cos(2\sin^{-1}x)}{\sqrt{1-x^{2}}}$
0%
$\dfrac{\cos(2\sin^{-1}x)}{\sqrt{1-x^{2}}}$
0%
$\dfrac{2\cos(2\cos^{-1}x)}{\sqrt{1-x^{2}}}$
0%
$-\dfrac{\cos(2\cos^{-1}x)}{\sqrt{1-x^{2}}}$

Explanation

We need to find derivative of

$\sin (2\sin^{-1}x)$

Apply the chain rule,

Let

$f=\sin(u)$

Therefore,

$\dfrac{d}{dx}\left(\sin \left(2\arcsin \left(x\right)\right)\right)=\dfrac{d}{du}\left(\sin \left(u\right)\right)\dfrac{d}{dx}\left(2\arcsin \left(x\right)\right)$

We know that

$\dfrac{d}{du}\left(\sin \left(u\right)\right)=\cos \left(u\right)$ and

$\dfrac{d}{dx}\left(2\arcsin \left(x\right)\right)=2\dfrac{d}{dx}\left(\arcsin \left(x\right)\right)=2\cdot \dfrac{1}{\sqrt{1-x^2}}$

So,

$\dfrac{d}{dx}\left(\sin \left(2\arcsin \left(x\right)\right)\right)$

$=\dfrac{d}{du}\left(\sin \left(u\right)\right)\dfrac{d}{dx}\left(2\arcsin \left(x\right)\right)$

$=\cos(u)\dfrac{2}{\sqrt{1-x^2}}=\cos \left(2\arcsin \left(x\right)\right)\dfrac{2}{\sqrt{1-x^2}}=\dfrac{2\cos \left(2\arcsin \left(x\right)\right)}{\sqrt{1-x^2}}$

Let

$y=\sin^{-1}x$ , then find

$(1-x^{2})y_{2}-xy_{1}$ .
Where

$y_{1}$ and

$y_{2}$ denote the first and second order derivatives respectively.

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

Explanation

Given, $y=\sin ^{ -1 }{ x }$

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

${ y }_{ 2 }=\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =-\dfrac { 1 }{ 2 } \dfrac { 1 }{ { (1-{ x }^{ 2 }) }^{ \dfrac { 3 }{ 2 } } } (-2x)=\dfrac { x }{ (1-{ x }^{ 2 })\sqrt { 1-{ x }^{ 2 } } }$

Thus the value of $(1-{ x }^{ 2 }){ y }_{ 2 }-x{ y }_{ 1 }$ is

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

$=\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } -\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } \\ =0$

So, option C is correct

Differentiate

$\cos^{-1}(4x^{3}-3x)$ w.r.t

$x$ .

0%
$\dfrac{3}{\sqrt{1-x^{2}}}$
0%
$-\dfrac{3}{\sqrt{1-x^{2}}}$
0%
$\dfrac{1}{\sqrt{1-x^{2}}}$
0%
$-\dfrac{1}{\sqrt{1-x}}$

Explanation

(Let the given function be

$y = f(x)$ .

$f(x)= \cos^{-1}(4x^{3}-3x)$

From the definition of first principle of differentiation,

$\dfrac{dy}{dx}= \lim_{x \rightarrow 0}\dfrac{f(x+\delta x)-f(x)}{\delta x}$

Now coming to the problem,

we know that

$\cos^{-1}(4x^3-3x)=3\cos^{-1}(x)$

$\dfrac{dy}{dx}= \lim_{\delta x \rightarrow 0}\dfrac{f(x+\delta x)-f(x)}{\delta x}$

$\dfrac{dy}{dx}= \lim_{\delta x \rightarrow 0}\dfrac{3{\cos}^{-1}(x+\delta x)-3{\cos}^{-1}x}{\delta x}$

$\dfrac{dy}{dx}= 3\lim_{\delta x \rightarrow 0}\dfrac{\cos^{-1}(x+\delta x)-{\cos}^{-1}(x)}{\delta x}$

We know derivative of

${]cos}^{-1}x$ is

$\dfrac {-1}{\sqrt {1-x^2}}$

Therefore,

$\dfrac{dy}{dx}= -\dfrac{3}{\sqrt {1-x^2}}$ .

Find derivative of

$\tan^{-1}\dfrac{\cos x}{1+\sin x}$ .

0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$
0%
$\dfrac{3}{2}$
0%
$-\dfrac{3}{2}$

Explanation

$\dfrac{d}{dx}\left(\arctan \left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right)$

$=\dfrac{d}{du}\left(\arctan \left(u\right)\right)\dfrac{d}{dx}\left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)$

We know that

$\dfrac{d}{du}\left(\arctan \left(u\right)\right)=\dfrac{1}{u^2+1}$ and

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

$=\dfrac{1}{-1-\sin \left(x\right)}$

So,

$\dfrac{d}{dx}\left(\arctan \left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right)=\dfrac{1}{u^2+1}\cdot \dfrac{1}{-1-\sin \left(x\right)}=\dfrac{1}{\left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)^2+1}\cdot \dfrac{1}{-1-\sin \left(x\right)}$

$=-\dfrac{\sin \left(x\right)+1}{\cos ^2\left(x\right)+\left(\sin \left(x\right)+1\right)^2}$

Therefore,

$f'(0)=-\dfrac{1}{2}$

$y=\dfrac{1}{2}(\sin^{-1}x)^{2}$ , then find

$(1-x^{2})y_{2}-xy_{1}$ .
Where

$y_{1}$ and

$y_{2}$ denote first and second derivatives of

$y$ respectively.

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

Explanation

Given

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

We have to find

$(1-x^2)y_2-xy_1$ where

$y_1,y_2$ denote the first and second derivatives

$y$ respectively.

$y_1=\dfrac{d}{dx}\left(\dfrac{1}{2}(sin^{-1}x)^2\right)$

$=\dfrac{1}{2} \cdot \dfrac{d}{dx}\left(sin^{-1} x\right)^2$

$=\dfrac{1}{2} \cdot 2 sin^{-1}x \cdot \dfrac{1}{\sqrt{1-x^2}}$

$y_1=\dfrac{sin^{-1} x}{\sqrt{1-x^2}}$

Now let us find

$y_2$

$y_2=\dfrac{d}{dx}y_1$

$=\dfrac{d}{dx}\left( \dfrac{sin^{-1}x}{\sqrt{1-x^2}}\right)$

$=\dfrac{\sqrt{1-x^2} \cdot \dfrac{d}{dx}(sin^{-1}x) - sin^{-1} x \cdot \dfrac{d}{dx} \left(\sqrt{1-x^2}\right)}{(\sqrt{1-x^2})^2}$

$=\dfrac{\sqrt{1-x^2} \cdot \dfrac{1}{\sqrt{1-x^2}} - sin^{-1} x \cdot \left(-\dfrac{x}{\sqrt{1-x^2}}\right)}{(\sqrt{1-x^2})^2}$

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

$\therefore y_2=\dfrac{\sqrt{1-x^2}+x sin^{-1}x}{(1-x^2)(\sqrt{1-x^2})}$

Now we have to find

$(1-x^2)y_2-xy_1$ .

$(1-x^2)y_2-xy_1=(1-x^2) \cdot \dfrac{\sqrt{1-x^2}+xsin^{-1}x}{(1-x^2)(\sqrt{1-x^2})}-x \cdot \dfrac{sin^{-1}x}{\sqrt{1-x^2}}$

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

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

$\therefore (1-x^2)y_2-xy_1=1$

The derivative of

$\sin^{-1}x$ with respect to

$\cos^{-1}\sqrt{1-x^2}$ is?

0%
$\displaystyle\frac{1}{\sqrt{1-x^2}}$
0%
$\cos^{-1}x$
0%
$1$
0%
$0$

Explanation

Assume that

$\theta=\sin^{-1}x$

$\therefore \sin\theta=x \implies \cos\theta=\sqrt{1-x^2}$

differentiating

$\sin\theta$ wrt

$x$ ,

$\cos\theta\dfrac{d\theta}{dx}=1$

$\therefore \dfrac{d\theta}{dx}=\dfrac{1}{\cos\theta}$

$\therefore \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{1-x^2}}$

Assume that

$\gamma =\cos^{-1}\sqrt{1-x^2}\implies \sin\gamma=x$

Also, differentiating

$\cos\gamma$ wrt

$x$ ,

$-\sin\gamma\dfrac{d\gamma}{dx}=\dfrac{1}{2}\dfrac{1}{\sqrt{1-x^2}}(-2x)$

$\therefore -x\dfrac{d\gamma}{dx}=\dfrac{1}{\sqrt{1-x^2}}(-x)$

$\therefore \dfrac{d\gamma}{dx}=\dfrac{1}{\sqrt{1-x^2}}$

Now, we require

$\dfrac{d\sin^{-1}x}{d\cos^{-1}\sqrt{1-x^2}}=\dfrac{d\theta}{d\gamma}$

$=\dfrac{d\theta}{dx}\times\dfrac{dx}{d\gamma}$

$=\dfrac{1}{\sqrt{1-x^2}}\times\dfrac{\sqrt{1-x^2}}{1}$

$=1$

This is the required answer.

$y=\sec^{-1}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\sin^{-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$ , then

$\dfrac{dy}{dx}=$

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

Explanation

Given,

$y={\sec}^{-1}\left (\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\sin}^{-1}\left (\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)$

We k.n.t

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

${\sec}^{-1}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}={\cos}^{-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$

Let

$m=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$

Thus

$y={\cos}^{-1}m+{\sin}^{-1}m=\dfrac{\pi}{2}$ (

$\because {\cos}^{-1}x+{\sin}^{-1}x=\dfrac{\pi }{2}$ )

$\Rightarrow \dfrac{dy}{dx}=\dfrac{d(\frac{\pi}{2})}{dx}=0$

$\Rightarrow \dfrac{dy}{dx}=0$

$y=\sin^{-1}(x^{2})$ then find

$\dfrac{dy}{dx}$ using first principle.

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

Explanation

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

$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^4}} \times 2x$

$=\dfrac{2x}{\sqrt{1-x^4}}$

$y=\cos^{-1}(\sqrt{x})$ , then find

$\dfrac{dy}{dx}$ using first principle.

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

Explanation

Given,

$f(x)=\cos^{-1}(\sqrt{x})$

Thus

$f(x+h)=\cos^{-1}(\sqrt{x+h})$

We know that

$f'(x)=\lim_{h\rightarrow0} \dfrac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\rightarrow0} \dfrac{\cos^{-1}(\sqrt{x+h})-\cos^{-1}(\sqrt x)}{h}$

$f'(x)=\lim_{h\rightarrow0}\dfrac{\left (\dfrac{\pi}{2}-\sin^{-1}(\sqrt{x+h})\right)-\left (\dfrac{\pi}{2}-\sin^{-1}(\sqrt{x})\right)}{h}$

$f'(x)=-\lim_{h\rightarrow0}\dfrac{\sin^{-1}(\sqrt{x+h})-\sin^{-1}(\sqrt{x})}{h}=-\lim_{h\rightarrow0}\dfrac{\sin^{-1}(\sqrt{x+h}(\sqrt{1-(\sqrt{x})^2})-\sqrt{x}(\sqrt{1-(\sqrt{x+h})^2})}{h}$

Multiply and divide by

$\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}- \sqrt{x}\sqrt{1-(\sqrt{x+h})^2}$

$=-\lim_{h\rightarrow0}\dfrac{\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}-\sqrt{x}\sqrt{1-(\sqrt{x+h})^2}}{h}$

Multiply and divide by

$\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h})^2}$

$=-\lim_{h\rightarrow0}\dfrac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h})^2})}=-\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$

Find derivative of

$\sin^{-1}(x^{2})$ using first principle.

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$\dfrac{2x}{\sqrt{1-x^{2}}}$
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$\dfrac{2x}{\sqrt{1-x}}$
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$\dfrac{2x}{\sqrt{1-x^{4}}}$
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$\dfrac{x}{\sqrt{1-x^{4}}}$

Explanation

It is given to us that

$\sin^{-1}(x^{2})$ = y, we can write

${x}^{2}$ =

$\sin(y)$ , so x =

$\sqrt{\sin(y)}$ ,

Now

$\dfrac{dx}{dy}$ =

$lim_{h \to 0}\dfrac{\sqrt{\sin(y+h)} - \sqrt{\sin(y)}}{h}$ (by using the first principle)

Now Rationalising the numerator by multiplying and dividing it by: (

$\sqrt{\sin(y+h)} + \sqrt{\sin(y)}$ )

After rationalisation we get:

$lim_{h \to 0} \dfrac{\sin(y+h) - \sin(y)}{h(\sqrt{\sin(y+h)} + \sqrt{\sin(y)})}$

Now using the formula:

$\sin(A) - \sin(B) = 2\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})$

We can write:-

$lim_{h \to 0} \dfrac{2\cos(y)\sin(\dfrac{h}{2})}{h(\sqrt{\sin(y+h)} + \sqrt{\sin(y)}}$ =

$lim_{h \to 0} \dfrac{\cos(y)}{\sqrt{\sin(y+h)} + \sqrt{\sin(y)}}$

$\dfrac{\sin(\dfrac{h}{2})}{\dfrac{h}{2}}$

Using the formula:

$lim_{h \to 0} \dfrac{\sin(h)}{h}$ = 1, also

$\sin(y+ h)$ is equal to

$\sin(y)$ when h is very very small

so after applying the limits we get

$\dfrac{dx}{dy}$ =

$\dfrac{\cos(y)}{2\sqrt{\sin(y)}}$

using

$\sin^{2}(a) + \cos^{2}(a) = 1$ , we can write

$\cos(a) = \sqrt{1-\sin^{2}(a)}$ and from the begining we can see that

$\sqrt{\sin(y)}$ = x

So finally we get

$\dfrac{dx}{dy}$ =

$\dfrac{\sqrt{1-x^{4}}}{2x}$

But we want the value of

$\dfrac{dy}{dx}$ which will be

$\dfrac{2x}{\sqrt{1-x^{4}}}$

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

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),

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

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

$x + y = \tan^{-1}y$ and

$y'' =f(y) y'$ then

$f(y) =$

0%
$\dfrac {1}{y(1+y^2)}$
0%
$\dfrac {3}{y(1+y^2)}$
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$\dfrac {2}{y(1+y^2)}$
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$\dfrac {-2}{y(1+y^2)}$

Explanation

$x+y =tan^{-1}y$

differentiating w.r.t. x

$1+ y' = \dfrac{1}{1+y^2} y'$

diffentiating once more,

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

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

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

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

$f(y) = \dfrac{-2}{y(1+y^2)}$

Let

$f$ be differentiable

$(x\epsilon R)$ , if

$f(2) = -2$ and

$f'(x) \geq 2$ for

$x\epsilon [1, 6]$ , then

0%
$f(6) < 6$
0%
$f(6) \geq 6$
0%
$f(6) = 5$
0%
$f(6) \leq 5$

Given

$\quad f(x)=\begin{cases} \log _{ a }{ { \left( a\left| \left[ x \right] +\left[ -x \right] \right| \right) }^{ x } } \left( \cfrac { { a }^{ \cfrac { 2 }{ \left( \cfrac { \left[ x \right] +\left[ -x \right] }{ \left| x \right| } \right) } -5 } }{ 3+{ a }^{ \cfrac { 1 }{ \left|x\right| } } } \right) \quad \text{for}\quad \left| x \right| \neq 0;a>1 \\ 0\quad \quad \quad \quad \quad \quad \quad \text{for}\quad x=0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \end{cases}$ ,
where

$\left[ . \right]$ represents the integral part function, then;

0%
$f$ is continuous but not differentiable at $x=0$
0%
$f$ is continuous and differentiable at $x=0$
0%
The differentiability of $f$ at $x=0$ depends on the value of $a$
0%
$f$ is continuous and differentiable at $x=0$ and for $a=e$ only

$y = \sin^{-1}\dfrac {1}{2}(\sqrt {1 + x} + \sqrt {1 - x})$ then

$y' =$

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

Explanation

$y=sin^{-1} (\dfrac{1}{2}(\sqrt{1+x}+\sqrt{1-x}))$

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

after solving,

$y' = \dfrac{-1}{2 \sqrt{1-x^2}}$

Let

$g: [1, 6]\rightarrow [0, \infty]$ be a real valued differentiable function satisfying

$g'(x) = \dfrac {2}{x + g(x)}$ and

$g(1) = 0$ , the maximum value of

$g$ cannot exceed.

0%
$ln\ 2$
0%
$ln\ 6$
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$6\ ln\ 2$
0%
$2\ ln\ 6$

Explanation

$f(x)=(x)$ for

$x\neq 0$

$=\bot$

$x=0$

$g(x)=\dfrac{2}{x+g(x)}$

$g(x)+x=\dfrac{2}{g(x)}\Rightarrow g(x)=\dfrac{2}{g(x)}-x$

$g(1)=\dfrac{2}{1}=2 \displaystyle\int g(x)=\int_{0}^{1}\dfrac{2}{g(x)}-x$

$=2\ln [g(x)]-\dfrac{x^{2}}{2}$

$=2\ln (2)-\dfrac{1}{2}$

Value of

$g(x)$ will never be greater than

$\ln 2$

1453900_768776_ans_c4db053aec334be3a37cb88b3cee47b0.png

Let

$f:(-1,1)\rightarrow R$ be the differentiable function with

$f(0)=-1$ and

$f'(0)=1$ .

$g(x)={ \left( f(2f(x)+2 \right) }^{ 2 }$ , then

$g'(0)=$

0%
$0$
0%
$-2$
0%
$4$
0%
$-4$

Explanation

Given,

$f(0)=-1, f'(0)=1$

Also given

$g(x)=f(2f(x)+2)^2$

Thus

$g'(x)=2f[(2f(x)+2)]2f'(x)$

$\Rightarrow g'(0)=2f[2f(0)+2]2f'(0)$

$\Rightarrow g'(0)=2[-1]2(1)=-4$

Determine the value of k for which the following function is continuous at

$x=3$ .

$f(x)=\dfrac{x^2-9}{x-3}, x \neq 3$

$f(x)=k, x=3$

0%
2
0%
4
0%
6
0%
8

Explanation

Since f(x) is continuous at

$x=3$ .

Therefore,

$\displaystyle\lim_{x\rightarrow 3} f(x) =f(3)$

$\displaystyle\lim_{x\rightarrow 3} f(x)=k$

$\displaystyle\lim_{x\rightarrow 3} \dfrac{x^2-9}{x-3}=k$

$\displaystyle\lim_{x\rightarrow 3} \dfrac{(x+3)(x-3)}{x-3}=k$

$\displaystyle\lim_{x\rightarrow 3} (x+3)=k$

$k=6$

Let

$f(x)=4$ and

$f'(x)=4$ , then

$\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { xf(2)-2f(x) }{ x-2 } }$

0%
$2$
0%
$-2$
0%
$-4$
0%
$4$

Explanation

$f\left( x \right) =4,f\left( x \right) =4$

$\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { xf\left( 2 \right) -2f\left( x \right) }{ x-2 } } =\cfrac { 0 }{ 0 }$ form

$=\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { 1.f\left( 2 \right) -2'f\left( x \right) }{ 1 } }$

$=f\left( 2 \right) -2f'\left( 2 \right)$

$=4-2(4)=-4$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 5 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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