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# Calculus/Differentiation

Question
Hello. I needed help with how to find dy/dx for :
y = sin^(-1)(a + bcosx / b + acosx)
I myself have solved the problem but I'm not sure if it's correct. Morever I have made a very lengthy solution. That's why I need your help. Thank You.

Now the derivative of sine^(-1)(f(x)) = f'(x)/(1 - f^2(x)).

In this case, f(x) = (a + bcosx)/(b + acosx), which is a quotient.

To find f'(x), it (hg' - gh')/h^2 where g = a + b*cos(x). and h = b + a*cos(x).
It can be seen that g' = -b*sin(x) and h' = -a*sin(x).

So take
g(x) = a + b*cos(x),
g'(x) = -b*sin(x)
h(x) = b + a*cos(x),
h'(x) =  -a*sin(x),

... but an even better way to do it is this.

Since sin(y) = (a + b*cos(x)) / (b + a*cos(x)), take the derivative.
Note the right is a quotient rule.

It gives cos(y)dy = ({[b+a*cos(x)][-b*sin(x)] - [a+b*cos(x)][-a*sin(x)]}/[b+a*cos(x)]^2)dx.

It is known that sin^2(y) + cos^2(y) = 1, so cos(y) = sqrt[1 - sin^2(y)].

Since we know that sin(y) =  (a + b*cos(x)) / (b + a*cos(x)), we can say that
cos(y) = sqrt[1 - ({a + b*cos(x)}/{b + a*cos(x)})^2].

Multiply both sides of the equation by dx/cos(y) and you've got the answer.
Yes, it is long.  So long I can't even express it simply.

That gives dy/dx = r(x)/s(x) where
r(x) = ({[b + a*cos(x)][-b*sin(x)] - [a + b*cos(x)][-a*sin(x)]}/[b + a*cos(x)]^2) and
s(x) = sqrt[1 - ({a + b*cos(x)}/{b + a*cos(x)})^2].

It should be noted that r(x) can be simplified to be
r(x) = sin(x)({[b + ab*cos(x)] - [a + ab*cos(x)]}/[b + a*cos(x)]^2).

Looking at this further, ab*cos(x) - ab*cos(x) = 0, so we're left with
r(x) = sin(x)([b-a]/[b + a*cos(x)]^2).

Now it can be said that
dy/dx = sin(x)([b-a]/[b + a*cos(x)]^2)/{sqrt[1 - ({a + b*cos(x)}/{b + a*cos(x)})^2]}.

Calculus

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