Find an equation of the line tangent to the curve y^2 + 4e^y = 4x + ln(x^2)
S sheltielov3r New member Joined Jan 31, 2010 Messages 1 Jan 31, 2010 #1 Find an equation of the line tangent to the curve y^2 + 4e^y = 4x + ln(x^2)
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jan 31, 2010 #2 At (1,0), m = 32 ⟹ y = 3x−32.\displaystyle At \ (1,0), \ m \ = \ \frac{3}{2} \ \implies \ y \ = \ \frac{3x-3}{2}.At (1,0), m = 23 ⟹ y = 23x−3. See graph.\displaystyle See \ graph.See graph. [attachment=0:230dd101]pqr.jpg[/attachment:230dd101] Attachments pqr.jpg 20 KB · Views: 152
At (1,0), m = 32 ⟹ y = 3x−32.\displaystyle At \ (1,0), \ m \ = \ \frac{3}{2} \ \implies \ y \ = \ \frac{3x-3}{2}.At (1,0), m = 23 ⟹ y = 23x−3. See graph.\displaystyle See \ graph.See graph. [attachment=0:230dd101]pqr.jpg[/attachment:230dd101]