Find the equation of the normal to the curve `y=(1+y)^y+sin^(-1)(sin^2x)a tx=0.`

Answer

`y + x - 1 =0`

Explanation

Given, `y = (1 + x)^(y) + sin^(-1) ( sin ^(2)x)`
Let `y = u + v,` where `u = (1 + x) ^(y), v = sin ^(-1) (sin ^(2 ) x)`.
On differentiating w.r.t. x, we get
`" " (dy)/(dx) = ( du)/( dx) + ( dv)/(dx) " " ` … (i)
Now, `" " u = (1 + x) ^(y)`
On taking logarithm both sides, we get
`log_e u = y log _e (1 + x)`
`rArr (1) / (u) ( du )/(dx) = ( y) /( 1+ x) + (dy )/( dx) { log_ e (1 + x) }`
`rArr (du ) /( dx) = (1 + x) ^(y) [ (y ) /(1+ x) + (dy )/(dx) log_e ( 1+ x)] " " ` ... (ii)
Again, `" " v = sin ^(-1) ( sin ^(2) x )`
`rArr " " sin v = sin ^(2) x `
`rArr cos v ""( dv) /(dx) = 2 sin x cos x `
`rArr " " (dv)/(dx) = (1) /( cos v ) ( 2sin x cos x )`
`rArr " " (dv) /( dx) = ( 2sin x cos x ) /( sqrt(1 - sin ^(2) v)) = ( 2sin x cos x ) /( sqrt(1 - sin ^(4)x )) " " ` ... (iii)
From Eq. (i),
`(dy)/(dx) = (1 + x ) ^(y) [ ( y) /( 1+ x) + (dy ) /(dx) log_ e ( 1+ x) ] + ( 2 sin x cos x ) /(sqrt(1- sin ^(4) x)) `
`rArr (dy )/(dx) = ( y(1+ x)^(y-1) + 2sin x cosx //sqrt(1 - sin ^(4) x )) /( 1- ( 1+ x) ^(y) log _ e (1 + x))`
At `x =0`,
`y = (1 + 0 ) ^(y) + sin ^(-1) sin (0) = 1 `
`therefore (dy)/(dx) = (1 (1 + 0 ) ^(1-1 ) + 2 sin 0 * cos 0 //sqrt((1 - sin ^(4) 0 )))/(1- ( 1+ 0 ) ^(1) log _e (1 + 0 )) `
`rArr (dy ) /(dx) = 1`
Again, the slope of the normal is
`" " m = - (1) /( dy//dx) = -1 `
Hence, the required equation of the normal is
`" " y - 1 = (-1 ) (x - 0)`
i.e., ` " " y + x - 1 =0`