The equation of the tangent line to the curve below at the point (1,1) is2x^2-y^4=1

The slope of the tangent line is equivalent to the value of the derivative of the curve at the given point.
derivative = 4x - 4y³
derivative at (1,1) = 4(1) - 4(1)³
derivative at (1,1) = 0

slope of tangent = 0

2 x² - y^4 = 1
y^4 = 2 x² - 1
y = (2x² - 1 )^(1/4)
y ` = 1/4 · ( 2 x² - 1 )^(-3/4) · 4 x = 
= x / (2x² - 1 )^(3/4) 
m = y`(1) = 1 / 1 = 1
m = 1
The equation of the tangent line:
y - y1 = m ( x - x1)
y - 1 = 1 · ( x - 1 )

The equation at the point ( 1, 1 ) is:  y = x

Ithink the answer would be knew