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X ∈ R,x = ±1 y ∈ R,y = 0 Explanation The denominator of y cannot be zero as this would make y undefined Equating the denominator Is there a horizontal asymptote in y = x21x HorizontalAnswer (1 of 3) Y=X^2 => \frac {dY}{dX} = 2X => slope of tangent of Y=X^2 at (1,1) = 2 (1)=2=m_1 Y^2=X => \frac {2YdY}{dX} =1 =>\frac {dY}{dX} = \frac {1}{2YLet us look into some examples to understand the above concept Example 1 Find the equation of the tangent to x 2 y 2 − 2x − 10y 1 = 0 at (− 3, 2) Solution Equation of tangent at (x 1, y 1)
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Verify that the functions y_1 (x) = c^{2}x \ and \ y_2(x) = xe^{2}x are the solutions to the differential equation y 4y 4y = 0 Verify that the functions y_1 (x) = \cos 2x \ and \ y_2 (x) = \sin 2x; Explanation ∙ xmtangent = dy dx at x = a y = x2 ⇒ dy dx = 2x x = 2 → dy dx = 4 using the pointslope form for equation of line y − 4 = 4(x − 2) ⇒ y = 4x − 4 Answer linkTour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site
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1)Find an equation of the tangent to the curve x = 2 ln t, y = t2 1 at the point (2, 2) by two methods (a) without eliminating the parameter y = ??Share It On Facebook Twitter Email 1 Answer 0 votes answered by Vikash Kumar (261kAtan2 (y, x) returns the angle θ between the ray to the point (x, y) and the positive x axis, confined to (−π, π Graph of over The function (from "2 argument arctangent ") is defined as the angle
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Incoming Term: tan^-1(x^2-y^2/x^2+y^2)=a, if tan^-1(x^2-y^2/x^2+y^2)=a, if tan^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=a then (dy)/(dx)=, if tan^-1(x^2-y^2/x^2+y^2)=e^a, cos^-1(x^2-y^2/x^2+y^2)=tan^-1a, tan^2(x+y)+cot^2(x+y)=1-2x-x^2, x^(2)y^(2)-tan^(-1)sqrt(x^(2)+y^(2))=cot^(-1)sqrt(x^(2)+y^(2)), (v) x^(2)y^(2)-tan^(-1)sqrt(x^(2)+y^(2))=cot^(-1)sqrt(x^(2)+y^(2)), y=(tan^(-1)x)^(2) (x^(2)+1)^(2)y_(2)+2x(x^(2)+1)y_(1)=2, if tan^ -1 (x^ 2 -y^ 2 /x^ 2 +y^ 2 )=a prove that dy/dx=x(1-tan)/y(1+tana),
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