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Q.

If y=axax..... then the value of dydx is

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a

y2x(1−ylogx)

b

y2logyx(1−ylogx)

c

y2logyx(1−ylogxlogy)

d

y2logyx(1+ylogxlogy)

answer is C.

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Detailed Solution

y=ax axax.....…⇒y=axylog⁡y=xylog⁡alog⁡(log⁡y)=ylog⁡x+log⁡(log⁡a)∣Diff   w.r. to ‘x’1log⁡y⋅1ydydx=yx+log⁡xdydx+odydx1ylog⁡y−log⁡x=yxdydx1−ylog⁡ylog⁡xylog⁡y=yx⇒dydx=y2log⁡yx[1−ylog⁡ylog⁡x]
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If y=axax..... then the value of dydx is