मराठी

Lf f(x) = cos-1 [1-(logx)21+(logx)2], then f' (e) = _______. -

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प्रश्न

lf f(x) = cos-1 `[(1 - (log x)^2)/(1 + (log x)^2)]`, then f' (e) = _______.

पर्याय

  • `1/"e"`

  • `2/"e"^2`

  • `2/"e"`

  • 1

MCQ

उत्तर

lf f(x) = cos-1 `[(1 - (log x)^2)/(1 + (log x)^2)]`, then f' (e) = `underline(1/"e")`.

Explanation:

We have

f(x) = `cos^-1 [(1 - (log x)^2)/(1 + (log x)^2)]`

f(x) = `2tan^-1 (log x) [because cos^-1 ((1 - x^2)/(1 + x^2)) = 2 tan^-1 x]`

f'(x) = `2/(1 + (log x)^2) xx 1/x`

f'(e) = `2/(1 + 1) xx 1/"e"`

`= 1/"e"`

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Higher Order Derivatives
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