Find equation of tangent n normal 2the given curve at indicated pt: y= x^3 (x to the power of 3) at (1,1)?

Question

This question is from Application of derivatives > Tangents and Normal.

Please explain it to me. And also tell me the basics of this topic.

Answer 1

y = x^3

=> dy/dx = 3x^2

=> slope of the tangent at (1, 1) is 3*(1)^2 = 3

=> equation of the tangent at (1, 1) is

y - 1 = 3(x - 1)

=> 3x - y = 2

Slope of the normal = - 1/3

=> equation of the normal through (1, 1) is

y - 1 = (-1/3) (x - 1)

=> x + 3y = 4.