If fx= log exx-2x+234 , x2-4≠ 0 then the value of dfdx at x = 3 is

f = log ex + log x-2x+234f = x + 34log x-2 - log x+2dfdx= 1 + 341x−2-1x+2Put x = 3= 1 + 341- 15= 1 + 3445= 85

If fx= log exx-2x+234 , x2-4≠ 0 then the value of dfdx at x = 3 is

f = log ex + log x-2x+234f = x + 34log x-2 - log x+2dfdx= 1 + 341x−2-1x+2Put x = 3= 1 + 341- 15= 1 + 3445= 85

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If $f (x) = \log [e^{x} {(\frac{x - 2}{x + 2})}^{\frac{3}{4}}], (x^{2} - 4 \neq 0)$ then the value of $\frac{d f}{d x}$ at x = 3 is

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$\frac{8}{5}$

$\frac{{8 e}^{3}}{5}$

answer is B.

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Detailed Solution

${f = log e}^{x} + log {(\frac{x-2}{x+2})}^{\frac{3}{4}}$

$f = x + \frac{3}{4} (log (x-2) - log (x+2))$

$\frac{df}{dx} = 1 + \frac{3}{4} (\frac{1}{x - 2} - \frac{1}{x+2})$

Put x = 3

$= 1 + \frac{3}{4} (1- \frac{1}{5})$

$= 1 + \frac{3}{4} (\frac{4}{5})$

$= \frac{8}{5}$

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If fx= log exx-2x+234 , x2-4≠ 0 then the value of dfdx at x = 3 is

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If $f (x) = \log [e^{x} {(\frac{x - 2}{x + 2})}^{\frac{3}{4}}], (x^{2} - 4 \neq 0)$ then the value of $\frac{d f}{d x}$ at x = 3 is