Mathematics 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): 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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