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(d^2x)/(dy^2) equals: ((d^2y)/(dx^2))^(-...

`(d^2x)/(dy^2)` equals: `((d^2y)/(dx^2))^(-1)` (b) `-((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-3)` `((d^2y)/(dx^2))((dy)/(dx))^(-2)` (d) `-((d^2y)/(dx^2))((dy)/(dx))^(-3)`

A

-1

B

1

C

log 2

D

`-log 2

Text Solution

Verified by Experts

The correct Answer is:
A
`x^(2x)-2x^(x)cot y-1=0" (i)"`
Now at x=1,
`1-2 cot y-1=0rArrcot y =0 rArry=(pi)/(2)`
Now differentiating (i) w.r.t. x, we get
`2x^(2x)(1+log x)-2[x^(x)(-cosec^(2)y)(dy)/(dx)+cot x^(2)(1+log x)]=0`
`"Now at "(1,pi//2)`,
`2(1+log 1)-2[1(-1)((dy)/(dx))_(((1,pi//2)))+0]=0`
`rArr" "2+2((dy)/(dx))_(((1,pi//2)))=0`
`((dy)/(dx))_(((1,pi//2)))=-1`
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