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Continuity and Differentiability Test - 21

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Continuity and Differentiability Test - 21
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  • Question 1
    1 / -0
    If $$y = x - x^{2}$$, then the derivative of $$y^{2} w.r.t. x^{2}$$ is
    Solution
    Given, $$y = x - x^{2}$$
    On differentiating w.r.t. $$x$$, we get
    $$\dfrac {dy}{dx} = 1 - 2x$$
    Now, $$\dfrac {d(y^{2})}{d(x^{2})} = \dfrac {\dfrac {d(y^{2})}{dx}}{d(x^{2})}{d(x)}$$
    $$= \dfrac {2y\dfrac {dy}{dx}}{2x} = 2y \dfrac {(1 -2x)}{2x}$$
    $$= \dfrac {2(x - x^{2})(1 - 2x)}{2x}$$
    $$= (1 - x)(1 - 2x)$$
    $$= 1 - 3x + 2x^{2}$$
    $$= 2x^{2} - 3x + 1$$.
  • Question 2
    1 / -0
    Find the derivative of $$\sin(2\sin^{-1}x)$$.
    Solution
    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, $$\dfrac{d}{dx}\left(\sin \left(2\arcsin \left(x\right)\right)\right)$$
    $$=\dfrac{d}{du}\left(\sin \left(u\right)\right)\dfrac{d}{dx}\left(2\arcsin \left(x\right)\right)$$ 
    $$=\cos(u)\dfrac{2}{\sqrt{1-x^2}}=\cos \left(2\arcsin \left(x\right)\right)\dfrac{2}{\sqrt{1-x^2}}=\dfrac{2\cos \left(2\arcsin \left(x\right)\right)}{\sqrt{1-x^2}}$$
  • Question 3
    1 / -0
    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.
    Solution

    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
  • Question 4
    1 / -0
    Differentiate $$\cos^{-1}(4x^{3}-3x)$$ w.r.t $$x$$.
    Solution
    (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, $$\dfrac{dy}{dx}= -\dfrac{3}{\sqrt {1-x^2}}$$.
  • Question 5
    1 / -0
    Find derivative of $$\tan^{-1}\dfrac{\cos x}{1+\sin x}$$.
    Solution
    $$\dfrac{d}{dx}\left(\arctan \left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right)$$
    $$=\dfrac{d}{du}\left(\arctan \left(u\right)\right)\dfrac{d}{dx}\left(\dfrac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)$$
    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, $$f'(0)=-\dfrac{1}{2}$$
  • Question 6
    1 / -0
    If $$y=\dfrac{1}{2}(\sin^{-1}x)^{2}$$, then find $$(1-x^{2})y_{2}-xy_{1}$$. 
    Where $$y_{1}$$ and $$y_{2}$$ denote first and second derivatives of $$y$$ respectively.
    Solution
    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 $$(1-x^2)y_2-xy_1$$.

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

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

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

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

  • Question 7
    1 / -0
    The derivative of $$\sin^{-1}x$$ with respect to $$\cos^{-1}\sqrt{1-x^2}$$ is?
    Solution
    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.
  • Question 8
    1 / -0
    If $$y=\sin^{-1}(x^{2})$$ then find $$\dfrac{dy}{dx}$$ using first principle.
    Solution
    $$y= \sin^{-1}(x^2)$$

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

    $$=\dfrac{2x}{\sqrt{1-x^4}}$$
  • Question 9
    1 / -0
    If $$y=\cos^{-1}(\sqrt{x})$$, then find $$\dfrac{dy}{dx}$$ using first principle.
    Solution
    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 $$\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h})^2}$$
    $$=-\lim_{h\rightarrow0}\dfrac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}\sqrt{1-(\sqrt{x})^2}+\sqrt{x}\sqrt{1-(\sqrt{x+h})^2})}=-\dfrac{1}{2\sqrt{x}\sqrt{1-x}}$$
  • Question 10
    1 / -0
    Find derivative of $$\sin^{-1}(x^{2})$$ using first principle.
    Solution
    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 $$\dfrac{2x}{\sqrt{1-x^{4}}}$$
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