Question

differentiate wrt x y=cos^-1((a-bsinx)/(a+bsinx))

Answer 1

implicitly find y'
from the equation:
\[\cos(y)=\frac{a-bsin(x)}{a+bsin(x)}\]

Answer 2

can u tell me the first step..

Answer 3

Take the derivative of both sides:
\[\frac{d}{dx}\cos(y)=\frac{d}{dx}\left[\frac{a-b\sin x}{a+b\sin x}\right]\]
Chain rule for the left, quotient for the right:
\[-\sin y\frac{dy}{dx}=\frac{(a+b\sin x)\dfrac{d}{dx}[a-b\sin x]-(a-b\sin x)\dfrac{d}{dx}[a+b\sin x]}{(a+b\sin x)^2}\]
Some derivatives:
\[-\sin y\frac{dy}{dx}=\frac{(a+b\sin x)(-b\cos x)-(a-b\sin x)(b\cos x)}{(a+b\sin x)^2}\]
Take out a common factor and simplify what you can
\[\begin{align*}-\sin y\frac{dy}{dx}&=\frac{-b\cos x\bigg(a+b\sin x+a-b\sin x\bigg)}{(a+b\sin x)^2}\\
\sin y\frac{dy}{dx}&=\frac{2ab\cos x}{(a+b\sin x)^2}
\end{align*}\]
and that's most of the work. Solve for \(\dfrac{dy}{dx}\).

Answer 4

\[y=\cos^{-1} \frac{ a-b \sin x }{ a+b \sin x }\]
\[\frac{ dy }{ dx }=-\frac{ 1 }{ \sqrt{1-\left( \frac{ a-b \sin x }{ a+b \sin x } \right)^2} }\frac{ d }{ dx }\left( \frac{ a-b \sin x }{ a+b \sin x } \right)\]
\[=-\frac{ 1 }{ \frac{\sqrt{a^2+b^2sin^2x+2 ab \sin x-(a^2+b^2sin^2x-2 ab \sin x)} }{ a+b \sin x } }*\frac{ \left( a+b \sin x \right)\left(- b \cos x \right)-\left( a-b \sin x \right)\left( b \cos x \right) }{ \left( a+b \sin x \right)^2 }\]
\[=\frac{ -a b \cos x-b^2\sin x \cos x-a b \cos x+b^2\sin x \cos x }{ \left( a+b \sin x \right)\sqrt{4 ab \sin x} }\]
\[=-\frac{ -2 ab \cos x }{ 2\sqrt{a b \sin x} \left( a+b \sin x \right)}=?\]