If y-Tan-13-2 log e x-1-6 log e x-Tan-1log ee-x2-log ee x2- then d 2 y-d x2 isA- 1B- 2C- 0D- -1

The correct option is C 0( 1 2 log_e^x/1+2 log_ex )+Tan^ 1 ( 3+2 log_ex/1 6 log_ex ) =Tan^ 1(1) Tan^ 1(2 log_ex)+Tan^ 1(3)+Tan^ 1(2 log_ex) =Tan^ 1(1)+Tan^ ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If y=Tan-13+2...

Question

If $y = T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) + T a n^{- 1} ⎛ ⎜ ⎝ \frac{l o g_{e} (\frac{e}{x^{2}})}{l o g_{e} (e x^{2})} ⎞ ⎟ ⎠, t h e n \frac{d^{2} y}{d x^{2}} i s$

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Solution

$(\frac{1 - 2 l o g_{e}^{x}}{1 + 2 l o g_{e} x}) + T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) = T a n^{- 1} (1) - T a n^{- 1} (2 l o g_{e} x) + T a n^{- 1} (3) + T a n^{- 1} (2 l o g_{e} x) = T a n^{- 1} (1) + T a n^{- 1} (3) ∴ \frac{d y}{d x} = 0$

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I f y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}), t h e n \frac{d^{2} y}{d x^{2}} i s

I f y = t a n^{- 1} (\frac{l o g_{e} (e / x^{2})}{l o g_{e} (e x^{2})}) + t a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}), t h e n \frac{d^{2} y}{d x^{2}} i s

i f y = T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 + 6 l o g_{e} x}) + T a n^{- 1} ⎛ ⎜ ⎝ \frac{l o g_{e} (\frac{e}{x^{2}})}{l o g_{3} (e x^{2})} ⎞ ⎟ ⎠, t h e n \frac{d^{2} y}{d x^{2}} i s

Q. If

y = {tan}^{- 1} ⎛ ⎝ \frac{log \frac{e}{x^{3}}}{log (e x^{3})} ⎞ ⎠ + {tan}^{- 1} (\frac{3 + 3 log x}{1 - 9 log (x)})

,then

\frac{d^{2} y}{d x^{2}} =

Q. If

y = {tan}^{- 1} (\frac{log e x^{- 2}}{log e x^{2}}) + {tan}^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

then

\frac{d^{2} y}{d x^{2}}

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If y=Tan−1(3+2logex1−6logex)+Tan−1⎛⎜⎝loge(ex2)loge(ex2)⎞⎟⎠, then d2ydx2 is

(1−2logxe1+2logex)+Tan−1(3+2logex1−6logex)=Tan−1(1)−Tan−1(2logex)+Tan−1(3)+Tan−1(2logex)=Tan−1(1)+Tan−1(3)∴dydx=0

If $y = T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) + T a n^{- 1} ⎛ ⎜ ⎝ \frac{l o g_{e} (\frac{e}{x^{2}})}{l o g_{e} (e x^{2})} ⎞ ⎟ ⎠, t h e n \frac{d^{2} y}{d x^{2}} i s$

$(\frac{1 - 2 l o g_{e}^{x}}{1 + 2 l o g_{e} x}) + T a n^{- 1} (\frac{3 + 2 l o g_{e} x}{1 - 6 l o g_{e} x}) = T a n^{- 1} (1) - T a n^{- 1} (2 l o g_{e} x) + T a n^{- 1} (3) + T a n^{- 1} (2 l o g_{e} x) = T a n^{- 1} (1) + T a n^{- 1} (3) ∴ \frac{d y}{d x} = 0$