Question
If xy.yx=16,then dydx at (2,2) is
Answer: Option A
:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
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:
A
xy.yx=16
∴logexy+logeyx=loge16
⇒ylogex+xlogey=4loge2
Now, differentiating both sides w.r.t.x
yx+logexdydx+xydydx+logey.1=0
∴dydx=−(logey+yx)(logex+xy)
∴dydx|(2,2)=−(loge2+1)(loge2+1)=−1
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