Calculus tangent to the curve

find the equation of the tangent to the curve 2x^2 − y^4 =1 at the point (–1,1)

the answer is y = −x

can you explain how to get it?

2x^2 - y^4 = 1

To find the equation of the tangent line at (-1, 1), you must use implicit differentiation.

4x - 4y^3 (dy/dx) = 0

Solve for dy/dx.

-4y^3(dy/dx) = -4x
dy/dx = (-4x)/[-4y^3]
dy/dx = x/y^3

So now, we can find the _slope_ at (-1, 1) by plugging these values into dy/dx. Use m to represent slope.

m = (-1)/(1)^3
m = -1

So now that we have slope m = -1, find the equation of the line with this slope that goes through (-1, 1).

(y2 - y1)/(x2 - x1) = m

(x1, y1) = (-1, 1)
(x2, y2) = (x, y)
m = -1

(y - 1)/(x - (-1)) = -1
(y - 1)/(x + 1) = -1
y - 1 = (-1)(x + 1)
y - 1 = -x - 1
y = -x