Find an equation of the line tangent to the curve y − 3x^2y^2 = cos x at the point (0, 1)

Answer 1

So every line requires a slope and a point.

The slope of the tangent line is the derivative when at the point (0, 1)

The line goes through (0,1), so therefore that point is on the line.

Let's get the derivative when at (0,1)

Use implicit differentiation:

dy/dx - [ 3x²(2ydy/dx) + y²(6x) ] = -sin x

Now plug in the point and simplify:

dy/dx - [ 3(0)²(2)(1)dy/dx + 1²(6)(0) ] = -sin (0)

dy/dx - 0 = 0

dy/dx = 0

The slope is 0 and our point is (0,1)

Use point-slope form:

y - 1 = 0(x - 0)

Simplifying gives y = 1

Answer 2

BY implicit differentiation: y' - 6xy² - 6x²yy' = - sinx;

At (0,1): y' = 0.

Equation of tangent line: y = 1 + 0(x - 0); y = 1.