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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives1 Mark
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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