Question

What is the equation of the tangent line of `f(x) =sin^2x-x^3` at `x=pi/4`?

Answer 1

`0.85055x+y-0.68355=0`. See tangent-inclusive Socratic graph.

Explanation:

Slope of the tangent is

`f'=2 sin x cos x-3 x^2=1-3/16(pi)^2=-.85055`, at `x = pi/4=0.7654`.

So, the equation to the tangent is

`y-(1-pi^3/64)=-0.85055(x-0.7654)`, giving

`0.85055x+y-0.68355=0`

graph{((sin x)^2-x^3-y)(.85x+y-.684) = 0 [-2.5, 2.5, -1.25, 1.25]}

graph{((sin x)^2-x^3-y)(.85x+y-.684) = 0 [-2.5, 2.5, -1.25, 1.25]}