#1**+2 **

(1).....really reaching back into my memory banks here

I believe that the answer is y = tan x + C

Note that the slopes shown in the field are always positive...and the slope of the tangent is always positve (where defined).........also.....the other cliue is that near 0, the slope of the tan decreases on both sides of 0 and then increases dramatically away from 0 to pi/2 (and - pi/2)

Look at the graph here : https://www.desmos.com/calculator/uceau9xjx7

I hope that I haven't mis-lead you

CPhill Jan 8, 2021

#2**+2 **

(2) Again.....I'm not totally sure...but....

I think the answer is y = sec x

Look at the graph here : https://www.desmos.com/calculator/v7cuzaovi5

As we approach 0 from the left side....the slope is negative and on the other side of 0 the slope is positive

Also....the other clue is that on each side of the asymptotes on the secant, the slopes occur in "pairs" negative-negative and positive-positive....notice that this occurs in the slope field, as well

Hope this is correct !!!

CPhill Jan 8, 2021

#3**0 **

Questions like this are always confusing.

I think this is what it means.

** Sorry Chris, I only just realized that you had already given the same graph.

here is the graph of y=tan x

**Can you see that the gradient (which is the derivative) is always positive?**

As the graph approaches the asymptotes, the gradient becomes infinitely large.

In the middle of each section, the gradient appears to be about 1. It is steeper than that everywhere else.

This looks the same as one of your graphs.

Which one is that?

Melody Jan 9, 2021