What is the slope of the tangent line to the graph of y= tan^2x at x = pi/4
So the slope of the tangent line is the derivative of the function.
yup, got that. then plug in pi/4
I assume that is where the unit circle comes into play
y' = [(tanx)^2]' = 2 * (tanx) * (Sec^2x) yeah and now just plug in pi/4
which is what I would like an explanation of pls!
driving me up the wall
what exactly is it saying when the equation is tan(pi/4)?
Okay so you want the slope when x = pi/4 you are given an equation: y = tan^2(x) y' = 2 * tanx * Sec^2(x) y' ( pi/4) = 2 * tan (pi/4) * Sec^2(pi/4)
all good with that - thanks
it"s calculating what tan (pi/4) is that is bothering me
or sec (pi/4)
oh lol tan (pi/4) = Sin (pi/4) / Cos (pi/4) Sec (pi/4) = 1 / Cos (pi/4)
yup broke it down to that
where do locate pi/4 though?
Now plug in: Sin (pi/4) = sqrt(2)/2 Cos (pi/4) = sqrt(2)/2
pi/4 is in radians, pi = 180 degrees
pi / 4 radians= 180 / 4 = 45 degrees
I used the unit circle to get the points - but, do I need to memorize this by heart?
oh sweet! that really helped!
okay glad i helped..
I see how you got the degrees, my issue is how shld I know that at that point of 45 degrees that sqrt2/2 is the point?
I know pull the x value for cos and the y value for sin,
but, do I need to memorize that at 45 degrees the corresponding point is (sqrt2/2, sqrt2/2)?
No no, when you get this: y' = 2 * tanx * Sec^2(x) plug in pi/4 wherever you see x
and then you wull get the point
my problem is that we are not allowed calculators in the exam
hence, my rather ridiculous question
okay then you will have to memorize Sin and Cos values of: 0, pi/6, pi/4, pi/3, pi/2, pi
ok, cool! can I also use pythagoras theorem?
yea why not
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