Question

hii guyz....find the derivative of the given pic​

Answer 1

Step-by-step explanation:

$y = \cos^{ - 1} ( \frac{a + b \cos(x) }{b + a \cos(x) } )$

$\frac{dy}{dx} = \frac{d}{dx}( \frac{a + b \cos(x) }{b + a \cos(x) } ). \frac{ - 1}{ \sqrt{1 - (\frac{a + b \cos(x) }{b + a \cos(x) } )^{2} } }$

$\frac{dy}{dx} = \frac{ - b \sin(x)(b + a \cos(x) ) + a \sin(x)(a + b \cos(x))}{ {(b + a \cos(x) )}^{2} } . \frac{ - (b + a \cos(x)) }{ \sqrt{ {(b + a \cos(x)) }^{2} - {(a + b \cos(x)) }^{2} } }$

On simplifying, we get

$\frac{dy}{dx} = \frac{( {b}^{2} - {a}^{2}) \sin(x) }{b + a \cos(x) } . \frac{1}{ \sin(x) \sqrt{ {b}^{2} - {a}^{2} } }$

$\frac{dy}{dx} = \frac{ \sqrt{ {b}^{2} - {a}^{2} } }{b + a \cos(x) }$