How Do You Differentiate Trigonometric Functions?

Click For Summary

Homework Help Overview

The discussion revolves around differentiating various trigonometric functions and sketching their graphs. The functions include combinations of sine, cosine, and secant, with specific intervals provided for analysis.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivatives of the functions, with attempts to clarify and correct each other's work. Questions arise regarding the accuracy of the derivatives and the application of trigonometric identities.

Discussion Status

Some participants have provided feedback on the differentiation attempts, suggesting corrections and clarifications. There is ongoing exploration of the derivatives, particularly for the third function, and participants are actively engaging with each other's reasoning.

Contextual Notes

There are indications of confusion regarding the use of constants such as π in the derivatives, and participants are encouraged to revisit their calculations. The discussion also touches on the presentation of the problems, with one participant mentioning a warning related to formatting choices.

domyy
Messages
196
Reaction score
0

Homework Statement



1. f(x) = 5 sin (8∏x)

2. g(x) = 4∏ [ cos (3∏x) sin (3∏x)]

3. h(x) = cos [sec (5∏x)]

4. Sketch the graph of each function on the indicated interval, making use of relative extrema and points of inflection.

f(x) = 2sinx + sin2x ; [0,2∏]


The Attempt at a Solution



1. f(x) = 5 sin (8∏x)

f'(x) = 5 cos (8∏x) . (8∏)

f'(x) = 40∏ cos (8∏x)

2. g(x) = 4∏ [ cos (3∏x) sin (3∏x)]

g'(x) = 4∏ {[cos (3∏x)][(sin (3∏x)]' + [sin (3∏x)][cos (3∏x)]'}

g'(x) = 4∏ {[cos (3∏x)][cos (3∏x) . 3∏ ] + [ sin (3∏x)][ -sin (3∏x) . 3∏ ]}

g'(x) = 12∏ cos2 (3∏x) -12∏ sin2 (3∏x)

3. h(x) = cos [sec(5∏x)]

h'(x) = -sin [ sec(5∏x) . 5∏][ tan (5∏x) . 5∏]

h'(x) = -25∏ sec (5∏x) tan(5∏x)

4. Sketch the graph of each function on the indicated interval, making use of relative extrema and points of inflection.

f(x) = 2sinx + sin2x

f'(x) = 2cosx + cos2x . 2

f'(x) = 2cosx + 2cos(2x)

=> f'(x) = 0

2cosx + 2cos(2x) = 0

Now, how to proceed from here?
 
Last edited by a moderator:
Physics news on Phys.org
hi domyy! :smile:

1 is ok

2 is ok, except haven't you lost one of the ∏s ?

3 is almost completely wrong, have another go, writing it out more carefully at each step

4 now you need to use one of the standard trigonometric identities (cosA - cosB), all of which you should learn :wink:
 
:shy: Hi!

3. h(x) = cos [sec(5∏x)]

h'(x) = -sin [ sec(5∏x)][ tan (5∏x)] . 5∏

h'(x) = -5∏sin [ sec (5∏x) tan(5∏x)]

How about now?

What's the wrong with ∏ on nr 2 ? :/
 
Last edited by a moderator:
I received a warning for excessive use of colors ??

I thought it was actually better for whoever was reading..to separate each problem by color since I was posting more than one.

I didn't do it because I was trying to decorate my post.

My bad.
 
Last edited:
hi domyy! :smile:

(btw, i didn't complain about the colours, but i didn't like them)

3 is now ok, except the derivative of sec is tan2 :redface:

(and, for 2, 4∏*3∏ = 12∏2 :wink:)
 
Yes. Got it!
=)

Now, I'm going to work on nr. 4
 

Similar threads

  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
Replies
4
Views
2K
Replies
10
Views
2K