The equation of the tangent to the curve `y=(1+x)^(y)+sin^(-1)(sin^(2)x)` at `x=0` is

Options

A. `x-y+1=0`
B. `x+y+1=0`
C. `2x-y+1=0`
D. `x+2y+2=0`

Answer

A

Explanation

Given `y=(1+x)^(y)+sin^(-1)(sin^(2)x)`
At `x=0, y=1`
Let `y=u+v` where `u=(1+x)^(y)`
`implieslogu=ylog(1+x)`
`implies 1/u (du)/(dx)=y/(1+x)+log (1+x)(dy)/(dx)`
and `v=sin^(-1)(sin^(2)x)`
`implies (dv)/(dx)=1/(sqrt(1-sin^(2)x)) . 2 sin x cos x =(sin 2x)/(cos x)`
Now `(dy)/(dx)=(du)/(dx)+(dv)/(dx)`
`implies(dy)/(dx)=(1+x)^(y)[y/(1+x)+log(1+x)(dy)/(dx)]+(sin 2x)/(cos x)` M
`=(dy)/(dx)[1-(1+x)^(y)log(x+1)]=((1+x)^(y)y)/(1+x)+(sin 2x)/(cos x)`
At `(0,1) (dy)/(dx)|_((0,1))=1/(1+0)=1`
`:.` Equation of tangent is
`(y-1)=1(x-0)impliesx-y+1=0`