Question

What is the equation of the line tangent to `f(x)=x^2- xcosx` at `x=0`?

Answer 1

Y=-x

Explanation:

First find slope by getting f ' (x)= 2x-[cosx - x sin x] At x=0, f '(x)= -1. At x=0, f(x)=0. Hence the tangent passes through (0,0) and has a slope of -1. Equation of tangent would, therefore, be y=-x