If y -log a x -log x a -log x x -log a a then dy - dx -A- 1- x - xlog aB- log a- x - x -log aC- 1- xloga - xlogaD- 1- xlog a -log a - x log a 2

The correct option is D 1/x log a log a/x(log a)^2y=log_a x + log_x a + log_x x + log_a a =log x/log a + log a/log x+1+log a/log a dy/dx=1/log a1/x+log a ( 1 ...

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Byju's Answer

Standard XII

Mathematics

Logarithm of a Complex Number

If y =log a x...

Question

If $y = l o g_{a} x + l o g_{x} a + l o g_{x} x + l o g_{a} a$ then $\frac{d y}{d x} =$

\frac{1}{x} + x l o g a

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\frac{l o g a}{x} + \frac{x}{l o g a}

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\frac{1}{x l o g a} + x l o g a

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\frac{1}{x l o g a} - \frac{l o g a}{x (l o g a)^{2}}

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Solution

The correct option is D $\frac{1}{x l o g a} - \frac{l o g a}{x (l o g a)^{2}}$
$y = l o g_{a} x + l o g_{x} a + l o g_{x} x + l o g_{a} a$
$= \frac{l o g x}{l o g a} + \frac{l o g a}{l o g x} + 1 + \frac{l o g a}{l o g a}$
$\frac{d y}{d x} = \frac{1}{l o g a} \frac{1}{x} + l o g a (\frac{- 1}{(l o g x)^{2}}) \frac{1}{x} + 0 + 0$
$= \frac{1}{x l o g a} - \frac{l o g a}{x (l o g x)^{2}}$

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If y=logax+logxa+logxx+logaa then dydx=

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If $y = l o g_{a} x + l o g_{x} a + l o g_{x} x + l o g_{a} a$ then $\frac{d y}{d x} =$