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

[lakshmi]

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

 

[HallsofIvy]

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

 

[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.

Did that answer your question?

 

[lakshmi]

i tried it but i am not gettting simplest form

 

[himanshu121]

Do u mean Differential Eqn

 

[Gokul43201]

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

 

[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.

<br /> ln(y) = ln(a + b \cos x) + ln(b+a \cos x)<br />

Differentiate both sides wrt x...

<br /> \frac{1}{y}\frac{dy}{dx} = \frac{-b\sin x}{a+b\cos x} + \frac{-a}{b+a \cos x}<br />

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

 

[arildno]

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

 

[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.]

 

[Jamez]

isn't SinX/CosX=TanX?

 

[Brennen]

by golly gosh, jimminy cricket there is right.

 

[Brennen]

isn't SinX/CosX=TanX?

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

 

[maverick280857]

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

Yeah where does that come from in this question?

 

[Jamez]

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

 

[Brennen]

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

hey...doesnt 2+2=4?

 

[futb0l]

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&#039;(x)-f(x)g&#039;(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.

Did that answer your question?

This answer should be right.