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Differentiation question

2011 · Shift 0 · Q43
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Differentiation question

2011 · Shift 0 · Q43

JEE MainMathematicsDifferentiationMCQ+4 / −1
d2xdy2{{{d^2}x} \over {d{y^2}}}dy2d2x​ equals:
  1. A
    −(d2ydx2)−1(dydx)−3- {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y​)−1(dxdy​)−3
  2. B
    (d2ydx2)(dydx)−2{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}(dx2d2y​)(dxdy​)−2
  3. C
    −(d2ydx2)(dydx)−3- \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y​)(dxdy​)−3
  4. D
    (d2ydx2)−1{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}(dx2d2y​)−1
View written solutionFree

Correct answer: C

  1. Let p=dydx.p=\frac{dy}{dx}.p=dxdy​. We need to find d2xdy2=ddy(dxdy).\frac{d^2x}{dy^2}=\frac{d}{dy}\left(\frac{dx}{dy}\right).dy2d2x​=dyd​(dydx​).

  2. Since dxdy=(dydx)−1=p−1,\frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1}=p^{-1},dydx​=(dxdy​)−1=p−1, we differentiate with respect to yyy: d2xdy2=ddy(p−1).\frac{d^2x}{dy^2}=\frac{d}{dy}(p^{-1}).dy2d2x​=dyd​(p−1).

  3. Use chain rule: ddy(p−1)=ddx(p−1)⋅dxdy.\frac{d}{dy}(p^{-1})=\frac{d}{dx}(p^{-1})\cdot \frac{dx}{dy}.dyd​(p−1)=dxd​(p−1)⋅dydx​. Now, ddx(p−1)=−p−2dpdx.\frac{d}{dx}(p^{-1})=-p^{-2}\frac{dp}{dx}.dxd​(p−1)=−p−2dxdp​. Also, dpdx=d2ydx2,dxdy=p−1.\frac{dp}{dx}=\frac{d^2y}{dx^2}, \qquad \frac{dx}{dy}=p^{-1}.dxdp​=dx2d2y​,dydx​=p−1. So, d2xdy2=(−p−2d2ydx2)p−1.\frac{d^2x}{dy^2}=\left(-p^{-2}\frac{d^2y}{dx^2}\right)p^{-1}.dy2d2x​=(−p−2dx2d2y​)p−1.

  4. Therefore, d2xdy2=−d2ydx2 p−3=−(d2ydx2)(dydx)−3.\frac{d^2x}{dy^2}=-\frac{d^2y}{dx^2}\,p^{-3}=-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}.dy2d2x​=−dx2d2y​p−3=−(dx2d2y​)(dxdy​)−3.

  5. Compare with options:

  • A: incorrect
  • B: incorrect
  • C: matches exactly
  • D: incorrect

Hence the correct option is C.\boxed{C}.C​.

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