MCQ
If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equals to :
  • $\frac{\text{f}\ '\text{g}\ ''-\text{g}\ '\text{f}\ ''}{(\text{f}\ ')^3}$
  • B
    $\frac{\text{f}\ '\text{g}\ ''-\text{g}\ '\text{f}\ ''}{(\text{f}\ ')^2}$
  • C
    $\frac{\text{g}\ ''}{\text{f}\ ''}$
  • D
    $\frac{\text{f}\ ''\text{g}\ '-\text{g}\ ''\text{f}\ '}{(\text{g}\ ')^3}$

Answer

Correct option: A.
$\frac{\text{f}\ '\text{g}\ ''-\text{g}\ '\text{f}\ ''}{(\text{f}\ ')^3}$
$\text{x}=\text{f}(\text{t})$
$\text{y}=\text{g}(\text{t}),$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{g}\ '}{\text{f}\ '}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}$
$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)\frac{\text{dt}}{\text{dx}}$
$=\frac{\text{f}'\text{g}\ ''-\text{g}'\text{f}\ ''}{\text{f}\ '^2}\frac{1}{\text{f}'}$
$=\frac{\text{f}\ '\text{g}\ ''-\text{g}'\text{f}''}{(\text{f}\ ')^3}$

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