3.- To answer this question, we will use the quotient rule:
[tex](\frac{f(x)}{g(x)})^{\prime}=\frac{g(x)f^{\prime}(x)-f^{}(x)g^{\prime}(x)}{(g(x))^2}.[/tex]
Before applying the rule, we will determine each derivative to avoid confusion:
[tex]\begin{gathered} (\cos x)^{\prime}=-\sin x, \\ (1+\sin x)^{\prime}=(1)^{\prime}+(\sin x)^{\prime}=0+\cos x=\cos x. \end{gathered}[/tex]
Now, we substitute the above results in the quotient rule:
[tex]f^{\prime}(x)=\frac{(1+\sin x)(-\sin x)-(\cos x)(\cos x)}{(1+\sin x)^2}.[/tex]
Simplifying the above result, we get:
[tex]f^{\prime}(x)=\frac{-\sin x-\sin ^2x-\cos ^2x}{(1+\sin x)^2}.[/tex]
Recall that:
[tex]\sin ^2x+cos^2x=1.[/tex]
Therefore:
[tex]f^{\prime}(x)=\frac{-\sin x-1}{(1+\sin x)^2}=\frac{-(\sin x+1)}{(1+\sin x)^2}=-\frac{1}{1+\sin x}.[/tex]
Answer:
[tex]f^{\prime}(x)=-\frac{1}{1+\sin x}.[/tex]
