Explanation
Step -1: Differentiating with respect to x
Given, y=xtany
⇒tan y=yx→(1)
Now,
⇒dydx=1×tany+xsec2ydydx [ Using Product rule of derivative ]
⇒dydx[1−xsec2y]=tany
⇒dydx=tany1−xsec2y→(2)
Step -2: Find the value of sec2y.
We know, sec2y−tan2y=1
⇒sec2y=1+tan2y
Putting the value from eq(1),
⇒sec2y=1+y2x2 →(3)
Step -3: Prove the desired.
From eq(2),
⇒dydx=tany1−xsec2y
⇒dydx=tany1−x(1+y2x2)
⇒dydx =xtanyx−x2−y2
⇒dydx =yx−x2−y2 [From eq(1)]
∴ If y=xtany then, dydx = yx−x2−y2.
Hence, the correct answer is option B.
2x+2y=2x+y
Differentiating both sides
ln2.2x+ln2.2ydydx=ln2.2x+y(1+dydx)2x+2ydydx=2x+y(1+dydx)2x+2ydydx=2x+y+2x+ydydx(2y−2x+y)dydx=(2x+y−2x)dydx=2x+y−2x2y−2x+ydydx=2x(2y−1)2y(1−2x)dydx=2x−y(2y−11−2x)
So option C is correct.
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