trey5498 wrote:Find an equation of the tangent line of f(x) = sin^2(x) at 1/2
I know the steps to solve this. Find the derivative, place derivative into the the original equation to find x
What do you mean by "placing the derivative" (which is f'(x) = 2sin(x)cos(x)) "into the original equation" (which is f(x) = sin^2(x))? What will you be creating by composing f' and f?
trey5498 wrote:place X into derivative to find Y.
Do you mean "place the x-value of the point of tangency into f'(x) in order to find the slope at the point" (which is y', not y)? Or are you referring to plugging the x-value (not the "X"-value) into f(x) to find the y-value for the point of tangency?
trey5498 wrote:Finally put into Y intercept form ( y-y1=m(x+x1) ).
Are you required to leave the equation in this preliminary form, or are you perhaps expected to simplify to obtain slope-intercept form?
trey5498 wrote:Since I have never done this question with sin I am just not sure about it.
How would the trig function alter the process? About which step are you "unsure"?