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

A
-5y
B
5y
C
25y
D
-25y

Solution:

$\frac{d x}{d θ} = - \sin θ, \frac{d y}{d θ} = 5 c o s 5 θ \frac{d y}{d x} = \frac{- 5 \cos 5 θ}{\sin θ}, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{- 5 \cos 5 θ}{\sin θ}) = \frac{d}{d θ} (\frac{- 5 \cos 5 θ}{\sin θ}) \frac{d θ}{d x}$

If $x = \cos θ, y = \sin 5 θ t h e n (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} =$

Solution:

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