Q107 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

If \(y=f(x)\) is twice differentiable function such that at a point \(P, \frac{d y}{d x}=4, \frac{d^2 y}{d x^2}=-3\), then \(\left(\frac{d^2 x}{d y^2}\right)_P=\)

  • A
    \(\frac{64}{3}\)
  • B
    \(\frac{16}{3}\)
  • C
    \(\frac{3}{16}\)
  • Correct answer
    D
    \(\frac{3}{64}\)

Solution

\(\begin{array}{ll} \because \frac{d^2 x}{d y^2} =\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d x}{d y} \frac{d}{d x}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right) \\ =\left(\frac{d x}{d y}\right)\left(\frac{-\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^2}\right)=-\frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d y}{d x}\right)^3} \\ \therefore\left(\frac{d^2 x}{d y^2}\right)_p =-\frac{(-3)}{(4)^3}=\frac{3}{64} \end{array}\) Hence, option (4) is correct.

Hence, the correct answer is option: 'D'.

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