If y -tan -1log e - x 3-log ex 3-tan -13-3 log x -1-9 log x - then d 2 y - dx 2-A- 1B- 0C- log 2D- log 3

The correct option is B 0y=tan^ 1(loge logx^3log e+log(x^3))+tan^ 1(3+3log x1 9log(x)) y=tan^ 1(1 logx^31+log(x^3))+tan^ 1(3+3log x1 3·3log(x)) y=tan^ 11 tan^ ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If y =tan -1l...

Question

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

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Solution

The correct option is B $0$
$y = {tan}^{- 1} (\frac{log e - log x^{3}}{log e + log (x^{3})}) + {tan}^{- 1} (\frac{3 + 3 log x}{1 - 9 log (x)})$
$y = {tan}^{- 1} (\frac{1 - log x^{3}}{1 + log (x^{3})}) + {tan}^{- 1} (\frac{3 + 3 log x}{1 - 3 \cdot 3 log (x)})$
$y = {tan}^{- 1} 1 - {tan}^{- 1} 3 log x + {tan}^{- 1} 3 + {tan}^{- 1} 3 log x$
$y = {tan}^{- 1} 1 - {tan}^{- 1} 3$
$\frac{d y}{d x} = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} = 0$

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If y=tan−1⎛⎝logex3log(ex3)⎞⎠+tan−1(3+3logx1−9log(x)),then d2ydx2=

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