# Defferentiate this to simplest formy=a+bcosx/b+acosx

1. Jun 20, 2004

### lakshmi

can anybody defferentiate this to simplest form
y=a+bcosx/b+acosx

2. Jun 20, 2004

### HallsofIvy

Staff Emeritus
Excuse me? bcosx/b= cos x doesn't it?

3. Jun 20, 2004

### Zorodius

I'm pretty sure he means

$$y = \frac {a + b \cos x}{b + a \cos x}$$

The rule you want to use to differentiate this is

$${d\over dx} \left[{f(x)\over g(x)}\right]= {g(x)f'(x)-f(x)g'(x)\over [g(x)]^2}$$

This is called the "quotient rule". In your problem, f(x) = a + b cos x, and g(x) = b + a cos x. If a and b are constants, then f'(x) = -b sin x, and g'(x) = -a sin x. You can substitute those into the quotient rule and then simplify the result, and your problem is solved.

4. Jun 21, 2004

### lakshmi

i tried it but i am not gettting simplest form

5. Jun 21, 2004

### himanshu121

Do u mean Differential Eqn

6. Jun 21, 2004

### Gokul43201

Staff Emeritus
Except for a pair of $$~ab~sinx~cosx$$ terms that cancel off in the numerator, there is no other simplification to do.

7. Jul 2, 2004

### maverick280857

NOTE: Solution is wrong

Okay here's a second method:

$$y = \frac {a + b \cos x}{b + a \cos x}$$

Take log to the base e (ln) of both sides.

$$ln(y) = ln(a + b \cos x) + ln(b+a \cos x)$$

Differentiate both sides wrt x...

$$\frac{1}{y}\frac{dy}{dx} = \frac{-b\sin x}{a+b\cos x} + \frac{-a}{b+a \cos x}$$

Cross-multiply (or multiply both sides by y) and simplify...

This method looks more tedious but is less prone to errors...the product rule can cause problems if the functions are complicated. But then there is no golden rule to say that one method is superior over the other..

Hope that helps...

Cheers

Vivek

Last edited: Jul 3, 2004
8. Jul 2, 2004

### arildno

Maverick: You should have a minus sign between the two logarithms, not a plus sign.

9. Jul 3, 2004

### maverick280857

Oh yeah arildno...please pardon my mistake :-)

There's also a mistake in the second term...which I have corrected here.

Here's the correct portion:

$$ln(y) = ln(a + b \cos x) + ln(b+a \cos x)$$

$$\frac{1}{y}\frac{dy}{dx} = \frac{-b\sin x}{a+b\cos x} + \frac{a\sin x}{b+a \cos x}$$

Cheers
Vivek
[NOTE: My previous solution is WRONG. Sorry for the inconvenience.]

10. Jul 5, 2004

### Jamez

isn't SinX/CosX=TanX?

11. Jul 5, 2004

### Brennen

by golly gosh, jimminy cricket there is right.

12. Jul 6, 2004

### Brennen

come to think of it...y did u actually bring that into question?

13. Jul 7, 2004

### maverick280857

Yeah where does that come from in this question?

14. Jul 8, 2004

### Jamez

:'( i was bored ok, and i wanted to know if it was right... sorry...

15. Jul 8, 2004

### Brennen

well u were....but... yeah whatever it doesn't matter.

hey...doesnt 2+2=4?

16. Jul 10, 2004