How can I find the equation of the tangent to x4+16∗y4=32 at (2, 1)?

Answer 1

The slope of line tangent to x⁴ + 16y⁴ = 32 at (2, 1) can be attained by taking the derivative. In this case we use implicit differentiation.

Apply the derivative operator to both sides of the equation:

d(x⁴ + 16y⁴) = d(32)/dx

dx

4x³ + 64y³(dy/dx) = 0 (the chain rule applies so the derivative of the y expression must include dy/dx)

64y³(dy/dx) = -4x³

dy/dx = -x3/16y³

Then plug in the point (x = 2, y = 1) to get dy/dx = -8/16 = -1/2.

Then use the point-slope form of a line to build the equation of the tangent line:

(y - y₁) = m(x - x₁)

(y - 1) = -1/2(x - 2)

If you want to convert it into the y = mx + b format you can but it's not necessary.

Answer 2

The slope of the tangent is the value of the derivative at (2,1) and you need to use implicit differentiation to get that slope.

derivative: 4x3 + 64y³(dy/dx) = 0

So plug in (2,1) into this equation and solve for the value of the derivative

Then that value = (y-1)/(x-2) is the equation of the tangent.