If y -log e x -log e a -log e x -log e a then dy - dx -A- log a-x-x-log aB- 1-x log a-x log aC- 1-x-x log aD- 1-x log a-log a-xlog x2

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

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 1

If y =log e x...

Question

If $y = l o g_{e} x + l o g_{e} a + l o g_{e} x + l o g_{e} a$ 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 x)^{2}}

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Solution

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

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If y=logex+logea+logex+logea then dydx=

The correct option is D 1x log a−log ax(log x)2y=logax+logxa+logxx+logex+logaa=log xlog a+log alog x+1+log alog adydx=1log a1x+log a(−1(log x)2)1x+0+0=1x log a−log ax(log x)2

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