F y -Tan-13-2 log e x -1-6 log x -Tan-1log e e - x 2-log 3 ex 2- then d 2 y - d x 2 isA- 0B- 1C- -1D- 2

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

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

f y =Tan-13+2...

Question

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

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Solution

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

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Similar questions

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

Q. 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

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

y = {tan}^{- 1} ⎡ ⎢ ⎢ ⎣ \frac{log (\frac{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}} =

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+2loge x1+6loge x)+Tan−1⎛⎜⎝loge(ex2)log3(ex2)⎞⎟⎠,then d2ydx2 is

y=Tan−1(1−2loge x1+2loge x)+Tan−1(3+2loge x1−6loge x)=Tan−1(1)−Tan−1(2loge x)+Tan−1(3)+Tan−1(2logex)=Tan−1(1)+Tan−1(3)∴dydx=0

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

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