Let be the function implicitly determined by the equation
Find the slope (angular coefficient) of the line tangent to on , located at the first quadrant of the Cartesian plane.
Give your answer to 1 decimal place.
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Letting y = 1 on the equation, we get
x 2 + 2 x = e ⇔ x 2 + 2 x + 1 = e + 1 ⇔ x + 1 = ± e + 1 ⇔ x = − 1 + e + 1
since we ought to have x > 0 , y > 0 so the point is located on the first quadrant of the Cartesian plane.
Using implicit derivation techniques on the displayed equation, we get
2 x + 2 y + 2 x ⋅ d x d y + y 3 ⋅ d x d y = e y ⋅ d x d y ⇔ ⇔ d x d y = e y − 2 x − y 3 2 x + 2 y
Therefore, d x d y ( − 1 + e + 1 , 1 ) = e − 2 ⋅ ( − 1 + e + 1 ) − 3 2 ⋅ ( − 1 + e + 1 ) + 2 = e − 2 e + 1 − 1 2 e + 1 ≈ − 1 . 8