Question

find the equation of the tangent line to
y = tan^2(3x) at x= pi/4

Answer 1

\[y'=2\tan(x)\sec^2(x)\] then you get the slope if you plug in \(\frac{\pi}{4}\)

Answer 2

oh , so I do the derivative and then plug in the x they give me?

Answer 3

why are you doing the derivative for the user???
also it is wrong!!

Answer 4

yes it is wrong, i am sorry
i did not pay attention to the \(3x\) insides

Answer 5

the slope of this line is: \(\large\color{black}{ f\prime(\pi/4) }\)

@misty1212 stop doing the poster work please!

Answer 6

yes, take the derivative to get a formula for the slope, then plug in the number and you will get the slope

Answer 7

Okay, @ilovereyvis_x3 do you know how to obtain the derivative of this function?

Answer 8

some what I know that d/dx (tanx) = sex^2x but here tan is ^2 power , so no I do not

Answer 9

I am going to re-write it, and you will tell me if you know how to proceed. Show your wrk though, please.

\(\large\color{black}{ y = \tan^2(3x) }\)

is same as: \(\large\color{black}{ y = (~\tan(3x)~~)^2 }\)

Answer 10

is it y prime is (sex(3x) * 3 )^2

Answer 11

omg I mean sec

Answer 12

you are to:
1) apply the power rule to \(\large\color{black}{ (~\tan(3x)~~)^2 }\) as if \(\large\color{black}{ \tan(3x) }\) is just an x.

2) apply the chain rule to the inner part, \(\large\color{black}{ \tan(3x) }\)

2) apply the chain rule to the inner of the inner part, \(\large\color{black}{ 3x }\)

Answer 13

no
sec ^2 (3x) * 3 )^2

Answer 14

can you tell me what a derivative of tan(x) is?

Answer 15

sec^2x

Answer 16

yes,

Answer 17

tell me what is the derivative of x^2?

Answer 18

(power rule)

Answer 19

2x

Answer 20

yes, and the derivative of 3x is 3.

Answer 21

yes

Answer 22

now, based on these 3 things we will differentiate your function, okay, lets do it together.
\(\large\color{black}{ y=~(~\tan(3x)~~)^2 }\)

Answer 23

ok

Answer 24

\(\large\color{black}{ y'=2~(~\tan(3x)~~)^{2-1} \times \sec ^2(3x)\times3}\)

\(\large\color{black}{ y'=2~(~\tan(3x)~~)^{1} \times \sec ^2(3x)\times3}\)

\(\large\color{black}{ y'=2\tan(3x) \times \sec ^2(3x)\times3}\)

Answer 25

see how I am applying the chain rule twice?

Answer 26

so you do the power rule thingy and move the 2 to the front and then keep what they gave you which is tan(3x) times sec^2(3x) and then chain rule 3

Answer 27

yes

Answer 28

cant you multiply 2 and 3?

Answer 29

so you get 6tan(3x) * sec^2(3x) ?

Answer 30

so you can notice that I first applied the power rule to the inner argument, then \(\large\color{black}{ \sec ^2(3x)}\) was the chain for \(\large\color{black}{ (~\tan(3x)~~)^{2}}\).

and then \(\large\color{black}{ 3}\) is the chain rule for \(\large\color{black}{ 3x}\)
that the \(\large\color{black}{ \sec ^2(3x)}\) has.

Answer 31

yes, after we found the derivative we CAN multiply 2 and 3, but i was just shoign where the 3 and 2 are coming from.

Answer 32

\(\large\color{black}{ y'=2\tan(3x) \times \sec ^2(3x)\times3}\)
becomes,

\(\large\color{black}{ y'=6\tan(3x) \times \sec ^2(3x)}\)

Answer 33

will use a different notation, okay?
I'll say

\(\large\color{black}{ f'(x)=6\tan(3x) \times \sec ^2(3x)}\)

Answer 34

yeah that is what I got

Answer 35

cool:)

Answer 36

So, now you need to plug in pi/4 for x, (in the derivatve)

Answer 37

and now I plug in the x ?

Answer 38

ok

Answer 39

how do I do sec^2 on my calculator

Answer 40

I will use a calculator for this, because I am lazy. Understanding the concept is important, but not calculating big expressions.

See how I am plugging it in?
http://www.wolframalpha.com/input/?i=6tan%283%CF%80%2F4%29+sec%5E2%283%CF%80%2F4%29

Answer 41

yes

Answer 42

so it is -12?

Answer 43

yes

Answer 44

so you know that:
\(\large\color{black}{ f~'(\pi/4)=-12}\)
\(\large\color{black}{x=\pi/4}\)

you need to find the y of the point, knowing that \(\large\color{black}{x=\pi/4}\) plug in \(\large\color{black}{\pi/4}\) for x into f(x).

Answer 45

ok but what will be the equation of the tangent line

Answer 46

find the y-coordinate of the point

Answer 47

(as I said, plug in \(\large\color{black}{\pi/4}\) into the f(x) )

Answer 48

y prime is out slope and we have out x but how do I find the y cordinate

Answer 49

oh the one given

Answer 50

we are given that \(\large\color{black}{x=\pi/4}\),
so the y-coordinate is going to be \(\large\color{black}{f(\pi/4)}\) correct?

Answer 51

i got 1

Answer 52

yes 1 is correct.

Answer 53

okay, a point slope formula says:


\(\large\color{black}{y-y_1=m(x-x_1)}\)

Answer 54

y-1=-12(x-pi / 4 ) ?

Answer 55

yes just like this. now simplify it

Answer 56

how ???

Answer 57

expand the right side

Answer 58

don't I just leave it like that?

Answer 59

you want it in y-intercept form probably.

Answer 60

or at least I would think you want it in a form of y=mx+b, but I don't know what your teacher might want you to do.

Answer 61

it just says write the equation , so I think I don't need to simplify

Answer 62

oh no , my teacher does not ask for it that way

Answer 63

oh, so then you are done. And even if not, you showed knowledge as far as the calculus part of the problem goes.

Answer 64

I have another one similar to this one but it sounds more confusing and I do not know what to do

Answer 65

Sure. but make a new question for it please.

Answer 66

by the way thank you for all the help and yes sure I will to give you another medal

Answer 67

I don't need a medal but I might go offline at any second. THere are those who need me and they can call me at any second. You were a joy to work with, yw;)