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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives1 Mark
MCQ
If $\text{y}=\tan^{-1}\Big\{\frac{\log(\frac{\text{e}}{\text{x}})^2}{\log(\frac{\text{e}}{\text{x}})^2}\Big\}+\tan^{-1}\Big(\frac{3-2\log,\text{x}}{1-6\log,\text{x}}\Big)$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
  • A
    $2$
  • B
    $1$
  • ✓
    $0$
  • D
    $-1$
✓

Answer

Correct option: C.
$0$
$\text{y}=\tan^{-1}\Big\{\frac{\log(\frac{\text{e}}{\text{x}})^2}{\log(\frac{\text{e}}{\text{x}})^2}\Big\}+\tan^{-1}\Big(\frac{3-2\log,\text{x}}{1-6\log,\text{x}}\Big)$
$=\tan^{-1}\Big\{\frac{1-2\log_\text{e}\text{x}}{1+2\log_\text{e}\text{x}}\Big\}+\tan^{-1}\Big\{\frac{3+2\log_\text{e}\text{x}}{1-6\log_\text{e}\text{x}}\Big\}$
$=\tan^{-1}1-\tan^{-1}(2\log_\text{e}\text{x})+\tan^{-1}(3)+\tan^{-1}(2\log_\text{e}\text{x})$
$=\tan^{-1}+\tan^{-1}(3)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

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