Chapter 2 - The Derivative - 2.5 Derivatives of Trigonometric Functions - Exercises Set 2.5 - Page 151: 5

$f'(x)=\frac{5\sin x-5\cos x+1}{(5+\sin x)^2}$

$f(x)=\frac{5-cosx}{5+sinx}$ $f'(x)=\frac{\frac{d}{dx}(5-cosx)\times(5+sinx)-(5-cosx)\times\frac{d}{dx}(5+sinx)}{(5+sinx)^2}$ $f'(x)=\frac{\frac{d}{dx}(5-cosx)\times(5+sinx)-(5-cosx)\times\frac{d}{dx}(5+sinx)}{(5+sinx)^2}$ $f'(x)=\frac{{sinx}\times(5+sinx)-(5-cosx)\times{cosx}}{(5+sinx)^2}$ $f'(x)=\frac{5sinx+sin^2x-5cosx+cos^2x}{(5+sinx)^2}$ $f'(x)=\frac{5sinx-5cosx+(sin^2x+cos^2x)}{(5+sinx)^2}$ $f'(x)=\frac{5sinx-5cosx+1}{(5+sinx)^2}$