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Answer

1

Explanation

Given, ` xe^(xy) = y+ sin^(2) x` ….(i)
On putting x = 0 , we get
` 0*e^(0)=y+0`
` rArr" " y = 0`
On differentiating Eq. (i) both sides w.r.t. x, we get
`1*e^(xy) +x*e^(xy)(x*(dy)/(dx) + y)=(dy)/(dx) +2 sin x cos x `
On putting x = 0, y = 0, we get
`e^(0)+0(0+0)=[(dy)/(dx)]_("(0,0)")=+2 sin 0 cos 0 `
` rArr" "[(dy)/(dx)]_(("0,0)") = 1`