Question

differentiate y= (a-bcosx) / (a+bcosx)

Answer 1

explain this first then I will try to help

Answer 2

what do you mean by explaining it?

Answer 3

like take a guess on what you think it is first

Answer 4

with respect to x, correct/

Answer 5

ohhh! I have no idea :( I'm stuck... But do we use the quotient rule?

Answer 6

yes, with respect to x.

Answer 7

Yep! With respect to x.

Answer 8

Quotient rule: \(\large\color{black}{ \frac{\LARGE d}{\LARGE dx}\left(\begin{matrix} \frac{\LARGE f(x)}{\LARGE g(x)} \\ \end{matrix}\right) =\frac{\LARGE g(x)f'(x)-f(x)g'(x)}{\LARGE [g(x)]^2}}\)

Answer 9

Does this rule make sense to you?

Answer 10

Yeah, i understand the rule but i'm just confused on how to differentiate a-bcosx .

Answer 11

oh, that we can take care:

Answer 12

tel me the derivative of cos(x).

Answer 13

- sin x

Answer 14

yes, so for -bcos(x), the derivative would be (-b)(-sin x)

Answer 15

which is same as bsin(x)

Answer 16

and so is the derivative of: a-bcos(x)
is equivalent to: bsin(x)

(because the "a" doesn't change anything)

Answer 17

so we know that: (using d/dx notation for derivative)
~ d/dx a-bcos(x) = bsin(x)

Answer 18

can you differentiate:
a+bcos(x) ?

Answer 19

oh! I think i understand it. So a+bcox (x) = -b sin(x) ??

Answer 20

yes, d/dx a+bcos(x) = -b sin(x)

Answer 21

So you know:

d/dx [ a-bcos(x) ] = bsin(x)
d/dx [ a+bcos(x) ] = -bsin(x)

use quotient rule: \(\large\color{teal}{ \frac{\LARGE d}{\LARGE dx}\left(\begin{matrix} \frac{\LARGE f(x)}{\LARGE g(x)} \\ \end{matrix}\right) =\frac{\LARGE g(x)f'(x)-f(x)g'(x)}{\LARGE [g(x)]^2}}\)


fill it in:
\(\large\color{teal}{ \frac{\LARGE d}{\LARGE dx}\left(\begin{matrix} \frac{\LARGE a-b\cos(x)}{\LARGE a+b\cos(x)} \\ \end{matrix}\right) =\frac{\LARGE ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}{\LARGE }}\)

Answer 22

need some help with codes?

Answer 23

yeah hahaha! I don't know how to fill it out online..

Answer 24

you can draw it, (click "draw" below)

Answer 25

or use my code:
```
\(\large\color{royalblue}{ \frac{\LARGE d}{\LARGE dx}\left(\begin{matrix} \frac{\LARGE f(x)}{\LARGE g(x)} \\ \end{matrix}\right) =\frac{\LARGE g(x)f'(x)-f(x)g'(x)}{\LARGE [g(x)]^2} }\)
```

becomes,

\(\large\color{royalblue}{ \frac{\LARGE d}{\LARGE dx}\left(\begin{matrix} \frac{\LARGE f(x)}{\LARGE g(x)} \\ \end{matrix}\right) =\frac{\LARGE g(x)f'(x)-f(x)g'(x)}{\LARGE [g(x)]^2} }\)