# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 24

$$g'(z)=\frac{-\csc^2z+\sin z\csc^2z+\cot z \cos z}{3(1-\sin z)^2}$$

#### Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\cot x)'=-\csc^2 x$. Recall that $(\sin x)'=\cos x$. Since $g(z)=\frac{\cot z}{3-3\sin z}$, then using the quotient rule, we have $$g'(z)=\frac{(3-3\sin z)(\cot z)'-\cot z (3-3\sin z)'}{(3-3\sin z)^2} \\=\frac{(3-3\sin z)(-\csc^2z)-\cot z (-3\cos z)}{(3-3\sin z)^2}=\frac{-3\csc^2z+3\sin z\csc^2z+3\cot z \cos z}{(3-3\sin z)^2}\\ =\frac{3(-\csc^2z+\sin z\csc^2z+\cot z \cos z)}{9(1-\sin z)^2}\\ =\frac{-\csc^2z+\sin z\csc^2z+\cot z \cos z}{3(1-\sin z)^2}$$

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