Find an equation of a Tangent Line of sin^2(x)?

[trey5498]

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 (which I believe I can skip that part since I know x is 1/2), place X into derivative to find Y. Finally put into Y intercept form ( y-y1=m(x+x1) ). Since I have never done this question with sin I am just not sure about it. Any ideas?

[nona.m.nona]

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?

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?

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?

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"?

[mabasamunashe]

impute (1/2) for x in f'(x) to obtain the grad of the line .then again for x in y=f(x) to obtain the y value thus the point becomes p(1/2;sin^2 1/2)......and grad 2sin(1/2)cos(1/2)........!