Question

I'm stuck on this trig derivative! =(

Answer 1

what exactly are you stuc on?

Answer 2

I have no clue how to do the 2nd part.

Answer 3

well if the tangent line is horizontal, then the rate of change of the line is horizontal, right?

Answer 4

if the rate of change of a function is 0, than by definition its derivative is 0, right?

Answer 5

@jmacar89

Answer 6

Yes so f'(x)=0

Answer 7

and what's f'(x)?

Answer 8

the derivative (the slope)?

Answer 9

What did you enter for f'(x)?

Answer 10

(5 (5 cos(x)+1))/(cos(x)+5)^2

Answer 11

(5 (5 cos(x)+1))/(cos(x)+5)^2=0

Answer 12

btw, you might want to check the parenthesis there

Answer 13

So then (5 (5 cos(x)+1))=(cos(x)+5)^2

Answer 14

yup

Answer 15

*nope

Answer 16

0=(5 (5 cos(x)+1))

Answer 17

0*((cos(x)+5)^2)=0

Answer 18

Alright so now all I have is (5 (5 cos(x)+1))=0. What do I do with this?

Answer 19

Solve for x within the range given

Answer 20

Can you explain that further? What does solving for the range mean?

Answer 21

You'll want to find the solution to the equation first.

Answer 22

0=(5 (5 cos(x)+1))

Answer 23

In the interval [0,2pi]

Answer 24

Solve the equation that you have, and throw out all the solutions that aren't in [0,2pi]

Answer 25

I'm a dummy and don't really know what you're talking about. Can you show me an example?

Answer 26

Apply this rule
\[\frac {d}{dx} ({\frac{f(x)}{g(x)}}) = \frac{f'(x)g(x)-f(x)g'(x)}{{g(x)}^2}\]

Answer 27

Ok.

Answer 28

We have a function.

Answer 29

We want to find tangent lines.

Answer 30

By definition, the slope of a tangent line is its derivative at that point.

Answer 31

We know the slope of the tangent line is 0 because it is horizontal.

Answer 32

Thus we reduce the problem down to
"try to find the derivative of the function, and look at when it's derivative is 0"

Answer 33

They gave us something else though. x must be in the interval [0,2pi]. Thus we limit all solutions to between 0 and 2pi

Answer 34

@jmacar89 , do you follow?

Answer 35

I understood everything up to having the slope at 0. I know that we have an interval, but how do I solve the equation for values in the interval?

Answer 36

Ok, let me show you.

Answer 37

Say we have a function.