Question

For `f(x)=x^2sinx^3` what is the equation of the tangent line at `x=pi`?

Answer 1

In slope-intercept form it is `y = (2pisinpi^3 + 3xpi^4cospi^2)x-2pi^2sinpi^3 + 3xpi^5cospi^2 + pi^2 sinpi^3`

Explanation:

`f(x) = x^2sinx^3`

The point where `x=pi` has `y` coordinate: `f(pi) = pi^2 sinpi^3`

To find the slope of the tangent line, we'll need the derivative:
`f'(x) = 2xsinx^3+x^2(cosx^3)3x^2`

`= 2xsinx^3 + 3x^4cosx^2`

At the point where `x=pi`, the slope of the tangent line is:
`m=f'(pi) = 2pisinpi^3 + 3xpi^4cospi^2`

The equation of the line through the point `(pi, pi^2 sinpi^3)` with slope `m=2pisinpi^3 + 3xpi^4cospi^2` is

`y- pi^2 sinpi^3 = (2pisinpi^3 + 3xpi^4cospi^2)(x-pi)`

If you prefer slope-intercept form it is:

`y = (2pisinpi^3 + 3xpi^4cospi^2)x-2pi^2sinpi^3 + 3xpi^5cospi^2 + pi^2 sinpi^3`