Question

What is the derivative of

1. csc x / sec x
2. (cot x + csc x) * (tan x - sin x)

Answer 1

well csc x / sec x = (1/sin x)/(1/cos x) = cos x/sin x

Now use the product rule.

Answer 2

cos x / sin x = cot x

Answer 3

Simplify the other expression you have there before you try and differentiate, it will make it easier; i.e., write it out in terms of sin and cos

Answer 4

yes, that's right, but it doesn't help you with the derivative unless you already know the derivative of cot x.

Answer 5

oh -sin x / cos x is derivative then...

Answer 6

Whoops! I am much closer...give me two minutes to figure it out...simplifying helped a bunch!

Answer 7

Quotient rule time

Answer 8

Yup. Figured that one out...lemme try the next one myself. I've gotten about 30 homework problems done, but those two I couldn't figure out because I wasn't getting the answer in the back of the book!

Answer 9

(sin x)(-sin x) - (cos x)(cos x) / sin ^2 x

-(sin ^2 x + cos ^2 x) / sin ^2 x

-1 / sin^2 x = -csc ^2 x

Answer 10

correct

Answer 11

Any chance I can get some help with the next one as well?

For the derivative, I'm getting 1-cos x / sin^2 x when the answer should be -cos x - sin x

Answer 12

I simplified the answer by getting everything into sin and cos, but I'm messing up somewhere, just not seeing where...

Answer 13

f(x) = (cot x + csc x)(tan x - sin x)
= (cos x + 1 / sin x)(-1)
= -cos x - 1 / sin x
f'(x) = (sin x)(sin x) - (cos x)(-cos x - 1) / sin^2 x
= sin^2x - (-cos^2x - cos x) / sin ^2 x
= 1 - cos x / sin^2 x

Answer 14

No. cot x = cos x/sin x

So
(cot x + csc x) * (tan x - sin x)
= (cos x / sin x + 1 / cos x)(sin x/cos x - sin x)
= cos x.sin x/sin x.cos x + sin x/cos^2 x - cos x - sin x/cos x
= 1 + sin x/cos^2 x - cos x - sin x/cos x

Now diferentiate each term in turn.

Answer 15

Give me one minute to try it on paper. Yeah, I got cos x / sin x for cot x, I just skipped steps when I was making the denominator equal on each side.

Answer 16

BTW, coming out of this exercise, just as you the derivative of sin x and cos x, it's worth calculating and learning by heart the derivatives of tan x, cot x, sec x and cosec x.

Answer 17

cot x = cos x / sin x
tan x = sin x / cos x
sec x = 1 / cos x
csc x = 1 / sin x

Right?

Answer 18

Yes, those are the definitions. Now calculate all their derivatives, check them 3 times, and then learn them all by heart.

Answer 19

e.g., d/dx of tan x = sec^2 x

Answer 20

I'm having a really hard time simplifying the original problem.

(cos x / sin x + 1 / sin x) * (sin x / cos x - sin x)
1 - cos x + 1 / cos x - 1
- cos x + 1 / cos x
-cos x + sec x

That's nothing like what you got.

Answer 21

Hey JamesJ,

I think what you got was wrong...csc x = 1/sin x, not 1/cos x, right?

Answer 22

oh yes, sorry.