Question
If xcos y+ycos x=5. Then
Answer: Option C
:
C
xcos y+ycos x=5
⇒ecos y logex+ecos x logey=5
∴ecos y loge x{cosyx−loge x(sin y)dydx}+ecosx logey{cos xydydx−sin x logey}=0
Putx=y=1, (cos 1−0)+(cos 1dydx−0)=0
∴dydx=−1
or y′=−1
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:
C
xcos y+ycos x=5
⇒ecos y logex+ecos x logey=5
∴ecos y loge x{cosyx−loge x(sin y)dydx}+ecosx logey{cos xydydx−sin x logey}=0
Putx=y=1, (cos 1−0)+(cos 1dydx−0)=0
∴dydx=−1
or y′=−1
Was this answer helpful ?
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