CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 8 - MCQExams.com

Let $$\displaystyle y={ x }^{ x }$$
$$\displaystyle \Rightarrow \log { y } =x\log { x }$$
On differentiating w.r.t. $$x$$ , we get
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\frac { x }{ x } +\log { x }$$
$$\displaystyle \Rightarrow \frac { dy }{ dx } =y\left( 1+\log { x }  \right)$$
              $$=  { x }^{ x }\left( \log { e } +\log { x }  \right)$$
              $$\displaystyle ={ x }^{ x }\left( \log { ex }  \right)$$

$$x^{2} + y^{2} = t + \dfrac {1}{t}$$     ....Given

$$f(x) = \log \left (e^{x} \left (\dfrac {x - 2}{x + 2}\right )^{\dfrac {3}{4}}\right )$$

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

$$\dfrac{dx}{d\theta}=\dfrac{d}{dθ}\left(a\cos \left(θ\right)+a\ln \left(\tan \left(\dfrac{θ}{2}\right)\right)\right)$$ $$=\dfrac{d}{dθ}\left(a\cos \left(θ\right)\right)+a\dfrac{d}{dθ}\left(\ln \left(\tan \left(\dfrac{θ}{2}\right)\right)\right)$$

Given that,
$${ e }^{ x }\left( 1+x \right) \cdot \sec ^{ 2 }{ x{ e }^{ x } } dx=f\left( x \right) +$$ constant
Put $$x{ e }^{ x }=t$$ in LHS
$$\Rightarrow \left( { e }^{ x }+x{ e }^{ x } \right) dx=dt$$
$$\therefore$$ LHS $$=\sec ^{ 2 }{ t } dt$$
      $$=\tan { t } +$$ constant
$$\Rightarrow \tan { \left( x{ e }^{ x } \right)  } +$$ constant $$=f\left( x \right) +$$ constant
$$\Rightarrow f\left( x \right) =\tan { \left( x{ e }^{ x } \right)  }$$

$$y=\log x^2$$

Consider the following
$$\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx$$
$$=cos\theta cosx-sin\theta.sinx$$                ....where  $$\theta =cos^{-1}(\dfrac{2}{\sqrt{13}})=sin^{-1}(\dfrac{3}{\sqrt{13}})$$
$$=cos(\theta+x)$$
Hence
$$cos^{-1}[\dfrac{2}{\sqrt{13}}cosx-\dfrac{3}{\sqrt{13}}sinx]$$
$$=cos^{-1}[cos(\theta+x)]$$
$$=\theta+x$$
$$=cos^{-1}(\dfrac{2}{\sqrt{13}})+x$$
Hence
$$f(x)=cos^{-1}(\dfrac{2}{\sqrt{13}})+x$$

$$f'(x)=1$$
Hence $$f'(x)$$ is a constant function.
Therefore
$$f'(0.5)=1$$ .

Given : $$f\left( x \right) ={ \left( x-1 \right)  }^{ 4 }{ \left( x-2 \right)  }^{ n } , n \in N$$

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

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

We have,   $$g(x)=[f(x)]^2+[f'(x)]^2$$

$$y= \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) + \tan^{-1}\left (\dfrac {1}{1 + (x + 1)(x + 2)}\right ) +\\ \quad\tan^{-1} \left (\dfrac {1}{1 + (x + 2)(x + 3)}\right ) + ..... + \tan^{-1} \left (\dfrac {1}{1 + (x + n - 1) (x + n)}\right )$$

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

Solution:

Solution:

$$\sec\left( \dfrac { x+y }{ x-y }  \right) =a$$ $$\longrightarrow (1)$$

We have,
$$y=\displaystyle\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+......\infty}}}}$$
$$\Rightarrow \displaystyle y^2=x+\sqrt{y+\sqrt{x+\sqrt{y+........\infty}}}$$
$$\Rightarrow y^2=x+\sqrt{y+y}$$ ...... [From the definition of $$y$$ ]

We have $${ 2 }^{ x }+{ 2 }^{ y }={ 2 }^{ x+y }$$

$$\textbf{Step -1: Differentiating with respect to x }$$

                  $$\text{Given,}$$   $${ y=x \tan y }$$

                  $${\Rightarrow\tan \ y=\dfrac{y}{x} \qquad\rightarrow (1)}$$

                  $$\text{Now,}$$  

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=1\times\tan y+x\sec ^{2}y\dfrac{\mathrm{d} y}{\mathrm{d} x}$$             $$\textbf{[ Using Product rule of derivative ]}$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}[1-x\sec ^{2}y]=\tan y$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\tan y}{1-x\sec ^{2}y}\rightarrow (2)$$

$$\textbf{Step -2: Find the value of }$$ $$\mathbf{ sec ^{2}y.}$$

                  $$\text{We know, }$$   $$\sec ^{2}y-\tan ^{2}y=1$$

                  $$\Rightarrow\sec ^{2}y=1+\tan ^{2}y$$

                  $$\text{Putting the value from eq(1),}$$  

                  $$\Rightarrow\sec ^{2}y=1+\dfrac{y^{2}}{x^{2}}$$                 $$\qquad\rightarrow (3)$$

$$\textbf{Step -3: Prove the desired.}$$

                  $$\text{From eq(2),}$$  

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\tan y}{1-x\sec ^{2}y}$$

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

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}$$ $$=\dfrac{x\tan y}{x-x^{2}-y^{2}}$$

                  $$\Rightarrow\dfrac{\mathrm{d} y}{\mathrm{d} x}$$ $$=\dfrac{y}{x-x^{2}-y^{2}}$$                       $$\textbf{[From eq(1)]}$$  

                  $$\therefore\ \text{If }$$ $$y=x\tan y\text{ then, }\dfrac{\mathrm{d} y}{\mathrm{d} x} \text{ = } \dfrac{ y}{x-x^{2}-y^{2}}.$$

$$\textbf{Hence, the correct answer is option B.}$$

$$\log _{ 10 }{ ({ { x }^{ 2 }-{ y }^{ 2 } }/{ { x }^{ 2 }+{ y }^{ 2 }) } } =2\\ \Rightarrow \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } ={ 10 }^{ 2 }\\ \Rightarrow 99{ x }^{ 2 }=-101{ y }^{ 2 }$$

$$\sin { x } =\cfrac { 2t }{ 1+{ t }^{ 2 } }$$  
Differentiate both sides w.r.t. t
$$\cos { x } \cfrac { dx }{ dt } =\cfrac { \left( 1+{ t }^{ 2 } \right) \left( 2 \right) -2t\left( 0+2t \right)  }{ { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } }$$
$$\cos { x } \cfrac { dx }{ dt } =\cfrac { 2+2{ t }^{ 2 }-4{ t }^{ 2 } }{ { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } }$$
$$\cos { x } \cfrac { dx }{ dt } =\cfrac { 2-2{ t }^{ 2 } }{ { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } }$$
$$\cfrac { dx }{ dt } =\cfrac { 2\left( 1-{ t }^{ 2 } \right)  }{ { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } } \times \cfrac { 1+{ t }^{ 2 } }{ \sqrt { { \left( 1+{ t }^{ 2 } \right)  }^{ 2 }-4{ t }^{ 2 } }  } \left[ \because \cos { x } =\sqrt { 1-\sin ^{ 2 }{ x }  }  \right]$$
$$\cfrac { dx }{ dt } =\cfrac { 2\left( 1-{ t }^{ 2 } \right)  }{ { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } } \times \cfrac { 1+{ t }^{ 2 } }{ \sqrt { 1+{ t }^{ 4 }+2{ t }^{ 2 }-4{ t }^{ 2 } }  }$$
$$\cfrac { dx }{ dt } =\cfrac { 2\left( 1-{ t }^{ 2 } \right)  }{ 1+{ t }^{ 2 } } \times \cfrac { 1 }{ \sqrt { { \left( 1+{ t }^{ 2 } \right)  }^{ 2 } }  } =\cfrac { 2 }{ 1+{ t }^{ 2 } } \longrightarrow (1)$$
Now, $$\tan { y } =\cfrac { 2t }{ 1-{ t }^{ 2 } }$$
$$\sec ^{ 2 }{ y } \cfrac { dy }{ dt } =\cfrac { \left( 1-{ t }^{ 2 } \right) \left( 2 \right) -2t\left( 0-2t \right)  }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } } =\cfrac { 2+2{ t }^{ 2 } }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } }$$
$$\cfrac { dy }{ dt } =\cfrac { 2\left( 1+{ t }^{ 2 } \right)  }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } } \times \cfrac { 1 }{ \left[ 1+{ \left[ \cfrac { 2t }{ 1-{ t }^{ 2 } }  \right]  }^{ 2 } \right]  } \left[ \because \sec ^{ 2 }{ x } =\tan ^{ 2 }{ x } +1 \right]$$
$$\cfrac { dy }{ dt } =\cfrac { 2\left( 1+{ t }^{ 2 } \right)  }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } } \times \cfrac { { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 }+4{ t }^{ 2 } } =\cfrac { 2{ \left( 1-{ t }^{ 2 } \right)  } }{ { \left( 1-{ t }^{ 2 } \right)  }^{ 2 } }$$
$$=\cfrac { 2 }{ 1+{ t }^{ 2 } } \longrightarrow (2)$$
$$\cfrac { dy }{ dx } =\cfrac { { dy }/{ dt } }{ { dx }/{ dt } } =\cfrac { { 2 }/{ { \left( 1-{ t }^{ 2 } \right)  } } }{ { 2 }/{ { \left( 1-{ t }^{ 2 } \right)  } } }$$
$$=1$$

Given curve: $$y^2e^{xy}=9e^{-3}x^2$$

Here, $$x\sin (a + y) + \sin a \cos (a + y) = 0$$      ..... (i)
On differentiating w.r.t. $$x$$ , we get
$$\dfrac {d}{dx} [x\sin (a + y)] + \dfrac {d}{dx} [\sin a\cos (a + y)] = 0$$
$$\Rightarrow \left [x\dfrac {d}{dx}\left \{\sin (a + y)\right \} + \sin (a + y) \dfrac {d}{dx}(x)\right ] + \sin a\dfrac {d}{dx} \cos (a + y) = 0$$
(using product rule and chain rule)
$$\Rightarrow \left [x \cos (a + y)\dfrac {d}{dx}(a + y) + \sin (a + y)\right ] + \sin a\left [-\sin (a + y) \dfrac {d}{dx} (a + y)\right ] = 0$$
$$\Rightarrow \left [x \cos (a + y) \left (0 + \dfrac {dy}{dx}\right ) + \sin (a + y)\right ] - \sin a \sin (a + y) \left (0 + \dfrac {dy}{dx}\right ] = 0$$
$$\Rightarrow x\cos (a + y) \dfrac {dy}{dx} + \sin (a + y) - \sin a \sin (a + y) \dfrac {dy}{dx} = 0$$
$$\Rightarrow \dfrac {dy}{dx} [x\cos (a + y) - \sin a \sin (a + y)] = -\sin (a + y)$$
$$\left [from Eq.(i), putting\ x = -\sin a \dfrac {\cos (a + y)}{\sin (a + y)}\right ]$$
$$\Rightarrow \dfrac {dy}{dx}\left [-\sin a\dfrac {\cos^{2}(a + y)}{\sin (a + y)} - \sin a \sin (a + y)\right ]$$
$$= -\sin (a + y)$$
$$\Rightarrow -\dfrac {dy}{dx}\left [\dfrac {\sin a \cos^{2} (a + y) + \sin a \sin^{2} (a + y)}{\sin (a + y)}\right ]$$
$$= -\sin (a + y)$$
$$\Rightarrow \dfrac {dy}{dx} = \sin (a + y) \left [\dfrac {\sin (a + y)}{\sin a \left \{\cos^{2} (a + y) +\sin^{2} (a + y)\right \}}\right ]$$
$$\Rightarrow \dfrac {dy}{dx} = \dfrac {\sin^{2}(a + y)}{\sin a} (\because \sin^{2}\theta + \cos^{2}\theta = 1)$$

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

Differentiating both sides

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

So option $$C$$ is correct.

Given, $$y = \dfrac {1}{1 + x + x^{2}} \Rightarrow 1 + x + x^{2} = \dfrac {1}{y}$$
On differentiating w.r.t. $$x$$ , we get
$$0 + 1 + 2x = \dfrac {1}{y^{2}} \dfrac {dy}{dx}$$
$$\Rightarrow \dfrac {dy}{dx} = -(1 + 2x)y^{2}$$ .

Given, $$u = \tan^{-1} \left \{\dfrac {\sqrt {1 - x^{2}} - 1}{x}\right \}$$
and $$v = \sin^{-1} x$$
Put $$x = \sin \theta \Rightarrow \theta = \sin^{-1}x$$ , then
$$u = \tan^{-1} \left \{\dfrac {\cos \theta - 1}{\sin \theta}\right \} (\because 1 - \cos^{2} \theta = \sin^{2} \theta )$$
$$= -\tan^{-1}\left \{\dfrac {1 - \cos \theta}{\sin \theta}\right \}$$
$$= -\tan^{-1} \left \{\dfrac {2\sin^{2} \theta /2}{2\sin \theta/2 \cdot \cos \theta /2}\right \}$$
$$= -\tan^{-1} \left (\tan \dfrac {\theta}{2}\right )$$
$$= -\dfrac {1}{2} \theta = -\dfrac {1}{2} \sin^{-1}x$$
Therefore, $$\dfrac {du}{dx} = -\dfrac {1}{2\sqrt {1 - x^{2}}}$$
and $$\dfrac {dv}{dx} = \dfrac {1}{\sqrt {1 - x^{2}}}$$
Thus $$\dfrac {du}{dv} = \dfrac {du}{dx}\times \dfrac {dx}{dv} = -\dfrac {1}{2\sqrt {1 - x^{2}}} \times \sqrt {1 - x^{2}}$$
$$= -\dfrac {1}{2}$$