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If R=([1+((dy)/(dx))^2]^(-3//2))/((d^2y)...

If `R=([1+((dy)/(dx))^2]^(-3//2))/((d^2y)/(dx2))` , then`R^(2//3)` can be put in the form of `1/(((d^2y)/(dx^2))^(2//3))+1/(((d^2x)/(dy^2))^(2//3))` b. `1/(((d^2y)/(dx^2))^(2//3))-1/(((d^2x)/(dy^2))^(2//3))` c. `2/(((d^2y)/(dx^2))^(2//3))+2/(((d^2x)/(dy^2))^(2//3))` d. `1/(((d^2y)/(dx^2))^(2//3))dot1/(((d^2x)/(dy^2))^(2//3))`

A

`(1)/(((d^(2)y)/(dx^(2)))^(2//3))+(1)/(((d^(2)y)/(dy^(2)))^(2//3))`

B

`(1)/(((d^(2)y)/(dx^(2)))^(2//3))+(1)/(((d^(2)y)/(dy^(2)))^(2//3))`

C

`(2)/(((d^(2)y)/(dx^(2)))^(2//3))+(2)/(((d^(2)y)/(dy^(2)))^(2//3))`

D

`(1)/(((d^(2)y)/(dx^(2)))^(2//3)).(1)/(((d^(2)y)/(dy^(2)))^(2//3))`

Text Solution

Verified by Experts

The correct Answer is:
A
Now `(dx)/(dy)=(1)/(((dy)/(dx)))`
`therefore" "(d^(2)x)/(dy^(2))=(1)/(dy)(1)/(((dy)/(dx)))`
`=(1)/(dx)(1)/(((dy)/(dx)))xx(dx)/(dy)`
`=-(1)/(((dy)/(dx))^(2))xx(d^(2)y)/(dx^(2))xx((dx)/(dy))`
`therefore" "(d^(2)x)/(dy^(2))=-(1)/(((dy)/(dx))^(3))xx(d^(2)y)/(dx^(2))`
`therefore" "((dy)/(dx))=((-(d^(2)y)/(dx^(2)))/(((d^(2)x)/(dy^(2)))))^(1//3)`
`1+((dy)/(dx))^(2)=1+((-(d^(2)y)/(dx^(2)))/((d^(2)x)/(dy^(2))))^(2//3)`
`"So, "(1+((dy)/(dx))^(2))/(((d^(2)y)/(dx^(2)))^(2//3))=(1)/(((d^(2)y)/(dx^(2)))^(2//3))+(((d^(2)y)/(dx^(2)))^(2//3))/(((d^(2)y)/(dx^(2)))^(2//3)((d^(2)x)/(dy^(2)))^(2//3))`
`=(1)/(((d^(2)y)/(dx^(2)))^(2//3))+(1)/(((d^(2)x)/(dy^(2)))^(2//3))`
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