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Question

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

A

1x+xloga
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B
log ax+xlog a
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C
1x log a+x log a
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D
1x log alog ax(log x)2
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Solution

The correct option is D 1x log alog ax(log x)2
y=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 alog ax(log x)2

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