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Question

If cos−1(x2−y2x2+y2)=k (a constant), then dydx=

A
yx
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B
xy
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C
x2y2
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D
y2x2
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Solution

The correct option is D yx


cos1x2y2x2+y2=k

x2y2x2+y2=cos(k)

On differentiating both sides, we get

(2x2ydydx)(x2+y2)(2x+2ydydx)(x2y2)(x2+y2)2=0

(2x3+2xy22x2ydydx2y3dydx)(2x32xy2+2x2ydydx2y3dydx)=0

4xy24yx2dydx=0

dydx=yx


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