Question

How do you find the equation of the line tangent to the graph of `y =sin^2x` at the point where x = pi/3?

Answer 1

Compute `y'(x)`, find the slope of the tangent line, `m =y'(pi/3)`, then use the point-slope form of a line and the point `(pi/3, y(pi/3))` to write the equation.

Explanation:

Compute y'(x):

The chain rule is `(d[f(g(x))])/dx = ((df)/(dg))((dg)/(dx))`

let `g(x) = sin(x)`, then `f(g) = g^2`, `(df)/(dg) = 2g` and `(dg)/dx = cos(x)`

`(d[sin²(x)])/dx = (2g)(cos(x))`

Reverse the substitution for g:

`(d[sin²(x)])/dx = (2sin(x))(cos(x))`

Use the identity `2sin(x)cos(x) = sin(2x)`

`(d[sin²(x)])/dx = sin(2x)`

The slope, `m = sin(2(pi/3))`

`m = sqrt3/2`

`y(pi/3) = sin²(pi/3)`

`y(pi/3) = 3/4`

Substitute these values into the point-slope form of a line:

`y - y_1 = m(x - x_1)`

`y - 3/4 = sqrt3/2(x - pi/3)`