Let `y` be an implicit function of `x` defined by `x^(2x)-2x^xcot y-1=0.` Then `y '(1)` equals: `1` b. `log2` c. `-log2` d. `-1`

Options

A. -1
B. 1
C. log 2
D. `-log 2

Explanation

`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`