The slope of the tangent to the curve (y-x^5)^2 = x (1+x^2)^2 at the point (1, 3) is

Question

This question belongs to the chapter TANGENT AND NORMAL

Answer 1

Differentiate wrt x:

2(y-x^5)(dy/dx-5x^4)=(1+x^2)^2+2x(1+x^2)2x⇒2(y-x^5)(dy/dx-5x^4)=(1+x^2)(1+5x^2)

Plug in (1,3):

2(3-1)(dy/dx-5)=(1+1)(1+5)

4dy/dx-20=12, 4dy/dx=32, dy/dx=8, the slope of the tangent.