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Find the derivative of:

$$f(x)=4 x^{3} \sin x \quad f(x)=4 x^{3}+2 x^{2}+\tan x \quad f(x)=\sqrt{x} \cot x \quad f(x)=4 \sec x \tan x$$

$$f(x)=\frac{1}{2 \cos x \cot x} \quad f(x)=\frac{1+\cot x}{2 x^{3}} \quad f(x)=\frac{1+\sec x}{2-\csc x}$$

$$D X[U V]=U^{\prime} V+V^{\prime} U$$

$$f(x)=4 x^{3} \sin x=4(3) x^{2} \sin x+4 x^{3} \cos x$$

$$f^{\prime}(x)=12 x^{2} \sin x+4 x^{3} \cos x$$

$$f(x)=4 x^{3}+2 x^{2}+\tan x \rightarrow f^{\prime}(x)=12 x^{2}+4 x+\sec ^{2} x$$

$$f(x)=\sqrt{x} \cot x \rightarrow f^{\prime}(x)=\frac{1}{2 \sqrt{x}} \cot x-\sqrt{x} \csc ^{2} x$$

$$f(x)=4 \sec x \tan x \longrightarrow f^{\prime}(x)=4 \sec x \tan x \tan x+4 \sec x \sec ^{2} x$$

$$f^{\prime}(x)=4 \sec x \tan ^{2} x+4 \sec ^{3} x$$

$$D x \left[\frac{u}{v}\right]=\frac{u^\prime v-u v^{\prime}}{v^{2}}$$

$$f(x)=\frac{1}{2 \cos x \cot x}=\frac{0-2(-\sin x) \cot x-\csc ^{2} x \cos x (1)}{(2 \cos x \cot x)^{2}}$$

$$f^{\prime}(x)=\frac{2 \sin x \cot x-{css}^{2} x \cos x}{4 \cos ^{2} x \cot ^{2} x}$$

$$f(x)=\frac{1+\cot x}{2 x^{3}}=\frac{-\csc ^{2} x\left(2 x^{3}\right)-6 x^{2}(1+\cot x)}{\left(2 x^{3}\right)^{2}}$$

$$f^{\prime}(x)=\frac{-2 x^{3} \csc ^{2} x-6 x^{2} \cot x-6 x^{2}}{4 x^{6}}$$

$$=\frac{2 x^{2}\left(-x \csc ^{2} x-3 \cot x-3\right)}{4 x^{6}}$$

$$f^{\prime}(x)=\frac{-x \csc ^{2} x-3 \cot x-3}{2 x^{4}}$$

$$f(x)=\frac{1+\sec x}{2-\csc x}=\frac{(\sec x \tan x)(2-\csc x)-(\csc x \cot x)(1+\sec x)}{(2-\csc x)^{2}}$$

$$f^{\prime}(x)=\frac{2 \sec x \tan x-\sec x \tan x \csc x-\csc x \cot x+\sec x \csc x \cot x}{4-4 \csc x+\csc ^{2} x}$$

Given $$f(x)=x^{2}+x \cos ^{2} x-1 .$$ Use the intermediate value
theorem to show that there is a real number $$C$$ between $$-\frac{\pi}{2}$$
and 0 such $$f^{\prime}(c)=0$$

$$f(x)=x^{2}+x \cos ^{2} x-1 \rightarrow f^{\prime}(x)=2 x+(1)\left(\cos ^{2} x\right)+(x)(2 \cos x (-\sin x))$$

$$f^{\prime}(x)=2 x+\cos ^{2} x-2 x \cos x \sin x$$

$$f^{\prime}(-\pi / 2)=2\left(-\frac{\pi}{2}\right)+\cos ^{2}\left(\frac{-\pi}{2}\right)-2\left(-\frac{\pi}{2}\right) \cos (-\pi / 2) \sin (-\pi / 2)=-\pi < 0$$

$$f^{\prime}(0)=2(0)+\cos ^{2}(0)-2(0) \cos (0) \sin (0)= 1 > 0$$

$$]c \in\left(-\frac{\pi}{2}, 0\right)$$ such that $$f^{\prime}(c)=0$$ "IVT"

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