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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 51–60 (part 6 of 11)

  1. Exercise 51

    If x=3sin⁡t−sin⁡3t,y=3cos⁡t−cos⁡3t\displaystyle x=3 \sin t-\sin 3 t, y=3 \cos t-\cos 3 t, find dydx\displaystyle \frac{d y}{d x} at t=π3\displaystyle t=\frac{\pi}{3}.

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    NCERT’s answer
    \(\displaystyle -\frac{1}{\sqrt{3}}\)
    \[\frac{dx}{dt} = 3\cos t - 3\cos 3t, \qquad \frac{dy}{dt} = -3\sin t + 3\sin 3t \]\[\frac{dy}{dx} = \frac{\sin 3t - \sin t}{\cos t - \cos 3t} \]\[\sin 3t - \sin t = 2\cos 2t\,\sin t, \qquad \cos t - \cos 3t = 2\sin 2t\,\sin t \]\[\frac{dy}{dx} = \frac{2\cos 2t\,\sin t}{2\sin 2t\,\sin t} = \cot 2t \]\[\left(\frac{dy}{dx}\right)_{t=\pi/3} = \cot\frac{2\pi}{3} = -\frac{1}{\sqrt{3}} \]Answer: \(\displaystyle -\dfrac{1}{\sqrt{3}}\)

  2. Exercise 52

    Differentiate xsin⁡x\displaystyle \frac{x}{\sin x} w.r.t. sin⁡x\displaystyle \sin x.

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    NCERT’s answer
    \(\displaystyle \frac{\tan x-x}{\sin ^2 x}\)
    Let \(\displaystyle u=\dfrac{x}{\sin x}\) and \(\displaystyle v=\sin x\).\[\frac{du}{dx} = \frac{\sin x - x\cos x}{\sin^2 x} \]\[\frac{dv}{dx} = \cos x \]\[\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\sin x - x\cos x}{\sin^2 x\,\cos x} \]Answer: \(\displaystyle \dfrac{\sin x - x\cos x}{\sin^2 x\cos x}\)

  3. Exercise 53

    Differentiate tan⁡−1(1+x2−1x)\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) w.r.t. tan⁡−1x\displaystyle \tan ^{-1} x when x≠0\displaystyle x \neq 0.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    Put \(\displaystyle x=\tan\theta\), so \(\displaystyle \theta=\tan^{-1}x\) with \(\displaystyle -\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}\), \(\displaystyle \theta\neq0\).\[u = \tan^{-1}\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \tan^{-1}\frac{\sec\theta-1}{\tan\theta} \]\[\frac{\sec\theta-1}{\tan\theta} = \frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2\dfrac{\theta}{2}}{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}} = \tan\frac{\theta}{2} \]\[u = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac12\tan^{-1}x \qquad \left(-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}\right) \]\[v = \tan^{-1}x \]\[\frac{du}{dv} = \frac{\dfrac{1}{2}\cdot\dfrac{1}{1+x^2}}{\dfrac{1}{1+x^2}} = \frac12 \]Answer: \(\displaystyle \dfrac{1}{2}\)
  4. Find \(\displaystyle \frac{dy}{dx}\) when \(\displaystyle x\) and \(\displaystyle y\) are connected by the relation given in each of the Exercises $\displaystyle 54$ to 57.

    Exercise 54

    sin⁡(xy)+xy=x2−y\displaystyle \sin (x y)+\frac{x}{y}=x^2-y

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    NCERT’s answer
    \(\displaystyle \frac{2 x y^2-y^3 \cos (x y)-y}{x y^2 \cos (x y)-x+y^2}\)
    Differentiate both sides with respect to \(\displaystyle x\).\[\cos(xy)\left(y + x\frac{dy}{dx}\right) + \frac{y - x\dfrac{dy}{dx}}{y^2} = 2x - \frac{dy}{dx} \]\[\frac{dy}{dx}\left[x\cos(xy) - \frac{x}{y^2} + 1\right] = 2x - y\cos(xy) - \frac{1}{y} \]Multiply by \(\displaystyle y^2\).\[\frac{dy}{dx}\left[xy^2\cos(xy) - x + y^2\right] = 2xy^2 - y^3\cos(xy) - y \]Answer: \(\displaystyle \dfrac{dy}{dx} = \dfrac{2xy^2 - y^3\cos(xy) - y}{xy^2\cos(xy) - x + y^2}\)

  5. Exercise 55

    sec⁡(x+y)=xy\displaystyle \sec (x+y)=x y

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    NCERT’s answer
    \(\displaystyle \frac{y-\sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y)-x}\)
    Differentiate both sides with respect to \(\displaystyle x\).\[\sec(x+y)\tan(x+y)\left(1 + \frac{dy}{dx}\right) = y + x\frac{dy}{dx} \]\[\frac{dy}{dx}\left[\sec(x+y)\tan(x+y) - x\right] = y - \sec(x+y)\tan(x+y) \]Answer: \(\displaystyle \dfrac{dy}{dx} = \dfrac{y - \sec(x+y)\tan(x+y)}{\sec(x+y)\tan(x+y) - x}\)

  6. Exercise 56

    tan⁡−1(x2+y2)=a\displaystyle \tan ^{-1}\left(x^2+y^2\right)=a

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    NCERT’s answer
    \(\displaystyle \frac{-x}{y}\)
    Differentiate both sides with respect to \(\displaystyle x\).\[\frac{1}{1+(x^2+y^2)^2}\left(2x + 2y\frac{dy}{dx}\right) = 0 \]The first factor is never zero, so\[2x + 2y\frac{dy}{dx} = 0 \]Answer: \(\displaystyle \dfrac{dy}{dx} = -\dfrac{x}{y}\)

  7. Exercise 57

    (x2+y2)2=xy\displaystyle \left(x^2+y^2\right)^2=x y

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    NCERT’s answer
    \(\displaystyle \frac{y-4 x^3-4 x y^2}{4 y x^2+4 y^3-x}\)
    Differentiate both sides with respect to \(\displaystyle x\).\[2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=y+x\frac{dy}{dx} \]\[4x(x^2+y^2)+4y(x^2+y^2)\frac{dy}{dx}=y+x\frac{dy}{dx} \]\[\frac{dy}{dx}\left[4y(x^2+y^2)-x\right]=y-4x(x^2+y^2) \]Answer: \(\displaystyle \dfrac{dy}{dx}=\dfrac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}\)

  8. Exercise 58

    If ax2+2hxy+by2+2gx+2fy+c=0\displaystyle a x^2+2 h x y+b y^2+2 g x+2 f y+c=0, then show that dydx⋅dxdy=1\displaystyle \frac{d y}{d x} \cdot \frac{d x}{d y}=1.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Differentiate with respect to \(\displaystyle x\).\[2ax+2h\left(y+x\frac{dy}{dx}\right)+2by\frac{dy}{dx}+2g+2f\frac{dy}{dx}=0 \]\[\frac{dy}{dx}=-\frac{ax+hy+g}{hx+by+f} \]Differentiate with respect to \(\displaystyle y\).\[2ax\frac{dx}{dy}+2h\left(x+y\frac{dx}{dy}\right)+2by+2g\frac{dx}{dy}+2f=0 \]\[\frac{dx}{dy}=-\frac{hx+by+f}{ax+hy+g} \]\[\frac{dy}{dx}\cdot\frac{dx}{dy}=\frac{ax+hy+g}{hx+by+f}\cdot\frac{hx+by+f}{ax+hy+g}=1 \]

  9. Exercise 59

    If x=exy\displaystyle x=e^{\frac{x}{y}}, prove that dydx=x−yxlog⁡x\displaystyle \frac{d y}{d x}=\frac{x-y}{x \log x}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Take logarithms of both sides.\[\log x=\frac{x}{y} \quad\Rightarrow\quad y=\frac{x}{\log x} \]\[\frac{dy}{dx}=\frac{\log x\cdot 1-x\cdot\frac1x}{(\log x)^2}=\frac{\log x-1}{(\log x)^2} \]\[\frac{x-y}{x\log x}=\frac{x-\dfrac{x}{\log x}}{x\log x}=\frac{x(\log x-1)}{x(\log x)^2}=\frac{\log x-1}{(\log x)^2} \]\[\frac{dy}{dx}=\frac{x-y}{x\log x} \]

  10. Exercise 60

    If yx=ey−x\displaystyle y^x=e^{y-x}, prove that dydx=(1+log⁡y)2log⁡y\displaystyle \frac{d y}{d x}=\frac{(1+\log y)^2}{\log y}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Take logarithms of both sides.\[x\log y=y-x \quad\Rightarrow\quad x(1+\log y)=y \quad\Rightarrow\quad \frac{x}{y}=\frac{1}{1+\log y} \]Differentiate \(\displaystyle x\log y=y-x\) with respect to \(\displaystyle x\).\[\log y+\frac{x}{y}\frac{dy}{dx}=\frac{dy}{dx}-1 \]\[\frac{dy}{dx}\left(1-\frac{x}{y}\right)=1+\log y \]\[1-\frac{x}{y}=1-\frac{1}{1+\log y}=\frac{\log y}{1+\log y} \]\[\frac{dy}{dx}=\frac{(1+\log y)^2}{\log y} \]