`"If "e^(sin(x^(2)+y^(3)))=tan""(y^(2))/(4)+sin^(-1)x," then y'(0) can be "`

Options

A. `(1)/(3sqrt(pi))`
B. `-(1)/(3sqrt(pi))`
C. `-(1)/(5sqrt(pi))`
D. `-(1)/(3sqrt(5pi))`

Explanation

`"We have "e^(sin(x^(2)+y^(2)))=tan""(y^(2))/(4)+sin^(-1)x" (1)"`
When x=0,
`e^(siny^(2))=tan""(y^(2))/(4)`
`rArr" "y=pmsqrt(x),pmsqrt(5x),...`
Now differentiating (1), w.r.t. x, we get
`e^(sin(x^(2)+y^(2)))[cos(x^(2)+y^(2))(2x+2y(dy)/(dx))]`
`=(2y)/(4)sec^(2)""(y^(2))/(4)(dy)/(dx)+(1)/(sqrt(1-x^(2)))`
Putting x= 0, we get
`rArr" "e^(siny^(2))[cos y^(2)(2y(dy)/(dx))]=(y)/(2)sec^(2)""(y^(2))/(4)(dy)/(dx)+1`
`"Now, at "x=0, y=sqrt(pi),(dy)/(dx)=(-1)/(3sqrt(pi))`
`"at "x=0,y=-sqrt(pi),(dy)/(dx)=(1)/(3sqrt(pi))`
`"at "x=0,y=sqrt(5pi),(dy)/(dx)=(-1)/(3sqrt(5pi))`
`"at "x=0, y=-sqrt(5pi),(dy)/(dx)=(1)/(3sqrt(5pi))`