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OpenStudy (anonymous):

What is the slope of the tangent line to the graph of y= tan^2x at x = pi/4

OpenStudy (anonymous):

4

OpenStudy (bahrom7893):

So the slope of the tangent line is the derivative of the function.

OpenStudy (anonymous):

yup, got that. then plug in pi/4

OpenStudy (anonymous):

I assume that is where the unit circle comes into play

OpenStudy (bahrom7893):

y' = [(tanx)^2]' = 2 * (tanx) * (Sec^2x) yeah and now just plug in pi/4

OpenStudy (anonymous):

which is what I would like an explanation of pls!

OpenStudy (anonymous):

driving me up the wall

OpenStudy (anonymous):

what exactly is it saying when the equation is tan(pi/4)?

OpenStudy (bahrom7893):

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)

OpenStudy (anonymous):

all good with that - thanks

OpenStudy (bahrom7893):

then what?

OpenStudy (anonymous):

it"s calculating what tan (pi/4) is that is bothering me

OpenStudy (anonymous):

or sec (pi/4)

OpenStudy (bahrom7893):

oh lol tan (pi/4) = Sin (pi/4) / Cos (pi/4) Sec (pi/4) = 1 / Cos (pi/4)

OpenStudy (anonymous):

yup broke it down to that

OpenStudy (anonymous):

where do locate pi/4 though?

OpenStudy (bahrom7893):

Now plug in: Sin (pi/4) = sqrt(2)/2 Cos (pi/4) = sqrt(2)/2

OpenStudy (bahrom7893):

pi/4 is in radians, pi = 180 degrees

OpenStudy (bahrom7893):

pi / 4 radians= 180 / 4 = 45 degrees

OpenStudy (anonymous):

I used the unit circle to get the points - but, do I need to memorize this by heart?

OpenStudy (anonymous):

oh sweet! that really helped!

OpenStudy (bahrom7893):

okay glad i helped..

OpenStudy (anonymous):

one sec

OpenStudy (anonymous):

:-)

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

I know pull the x value for cos and the y value for sin,

OpenStudy (anonymous):

but, do I need to memorize that at 45 degrees the corresponding point is (sqrt2/2, sqrt2/2)?

OpenStudy (bahrom7893):

No no, when you get this: y' = 2 * tanx * Sec^2(x) plug in pi/4 wherever you see x

OpenStudy (bahrom7893):

and then you wull get the point

OpenStudy (anonymous):

my problem is that we are not allowed calculators in the exam

OpenStudy (anonymous):

hence, my rather ridiculous question

OpenStudy (bahrom7893):

okay then you will have to memorize Sin and Cos values of: 0, pi/6, pi/4, pi/3, pi/2, pi

OpenStudy (anonymous):

ok, cool! can I also use pythagoras theorem?

OpenStudy (bahrom7893):

yea why not

OpenStudy (anonymous):

ok cool

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