यदि `x=2cos theta-cos 2theta` तथा `y=2sin theta-sin 2theta` तो `((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))` पर ज्ञात करें |

Explanation

`x=2cos theta-cos 2theta" "implies" "(dx)/(d theta)=-2sin theta+2sin 2 theta" "…(1)`
`y=2 sin theta-sin 2 theta" "implies" "(dy)/(d theta)=2 cos theta-2cos 2theta" "…(2)`
अब ` (dy)/(dx)=(dy//d theta)/(dx//d theta)=(cos theta-cos 2theta)/(sin 2theta-sin theta)`
`=(2 sin""(3 theta)/(2)sin""(theta)/(2))/(2 cos""(3 theta)/(2)sin""(theta)/(2))=tan""(3theta)/(2)`
पुन: x के सापेक्ष अवकलित (Differentiate) करने पर हमें मिलता है,
`(d^(2)y)/(dx^(2))=sec^(2)""(3theta)/(2).(3)/(2)(dtheta)/(dx)`
`=(3)/(2)sec^(2)""(3theta)/(2).(1)/(2(sin2theta-sin theta))" "[(1)" से"]`
`implies((d^(2)y)/(dx^(2)))_(theta=(pi)/(2))=(3)/(4)sec^(2)""(3pi)/(4).(1)/(sinpi-sin""(pi)/(2))`
`=(3)/(4)cdot2cdot(1)/(0-1)=-(3)/(2)`