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Q.

If  x=cosθ,y=sin5θ  then(1x2)d2ydx2xdydx=

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a

25y

b

-25y

c

-5y

d

5y

answer is D.

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Detailed Solution

dxdθ=-sinθ, dydθ=5cos5θ  dydx=-5cos5θsinθ,   d2ydx2=ddx-5cos5θsinθ=ddθ-5cos5θsinθdθdx

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If  x=cosθ, y=sin5θ  then(1−x2)d2ydx2−xdydx=