2 Find the derivative of each function. (a) $$ \mathrm{h}(t)=t an t $$ (b) $$ \mathrm{g}(x)=x^{3} \cos x $$ (c) $$ v( heta)=( heta+2)^{2} \sin heta $$ (d) $$ y=\frac{\sin x}{1+2 x-x^{2}} $$ (e) $$ z=\sin (2 heta+3)^{23} $$ (f) $$ p= an x \sec x $$

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2 Find the derivative of each function. (a) $$ \mathrm{h}(t)=t 	an t $$ (b) $$ \mathrm{g}(x)=x^{3} \cos x $$ (c) $$ v(	heta)=(	heta+2)^{2} \sin 	heta $$ (d) $$ y=\frac{\sin x}{1+2 x-x^{2}} $$ (e) $$ z=\sin (2 	heta+3)^{23} $$ (f) $$ p=	an x \sec x $$

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(a) For $$ h(t)=t \tan t $$, we use the product rule:
$$
h'(t)=\frac{d}{dt}[t]\cdot \tan t + t\cdot \frac{d}{dt}[\tan t]
$$
Since $$\frac{d}{dt}[t]=1$$ and $$\frac{d}{dt}[\tan t]=\sec^2 t$$, we have:
$$
h'(t)=\tan t + t \sec^2 t
$$

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(b) For $$ g(x)=x^3 \cos x $$, applying the product rule gives:
$$
g'(x)=\frac{d}{dx}[x^3]\cdot \cos x + x^3\cdot \frac{d}{dx}[\cos x]
$$
With $$\frac{d}{dx}[x^3]=3x^2$$ and $$\frac{d}{dx}[\cos x]=- \sin x$$,
$$
g'(x)=3x^2 \cos x - x^3 \sin x
$$

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(c) For $$ v(\theta)=(\theta+2)^2 \sin \theta $$, using the product rule:
$$
v'(\theta)=\frac{d}{d\theta}[(\theta+2)^2]\cdot \sin \theta + (\theta+2)^2\cdot \frac{d}{d\theta}[\sin \theta]
$$
The derivative of $$(\theta+2)^2$$ is $$2(\theta+2)$$ (by the chain rule) and $$\frac{d}{d\theta}[\sin \theta]=\cos \theta$$. Thus,
$$
v'(\theta)=2(\theta+2)\sin \theta+(\theta+2)^2\cos \theta
$$

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(d) For $$ y=\frac{\sin x}{1+2x-x^2} $$, we apply the quotient rule:
$$
y'=\frac{(1+2x-x^2)\frac{d}{dx}[\sin x]-\sin x\frac{d}{dx}[1+2x-x^2]}{(1+2x-x^2)^2}
$$
Here, $$\frac{d}{dx}[\sin x]=\cos x$$ and $$\frac{d}{dx}[1+2x-x^2]=2-2x$$. Therefore,
$$
y'=\frac{(1+2x-x^2)\cos x - \sin x(2-2x)}{(1+2x-x^2)^2}
$$

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(e) For $$ z=\sin\bigl((2\theta+3)^{23}\bigr) $$, we use the chain rule. Let
$$
u=(2\theta+3)^{23}
$$
Then $$ z=\sin u $$ with derivative:
$$
z'=\cos u\cdot u'
$$
Differentiate $$ u $$ using the chain rule:
$$
u'=23(2\theta+3)^{22}\cdot \frac{d}{d\theta}[2\theta+3]=23(2\theta+3)^{22}\cdot 2=46(2\theta+3)^{22}
$$
Thus,
$$
z' = 46(2\theta+3)^{22}\cos\bigl((2\theta+3)^{23}\bigr)
$$

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(f) For $$ p=\tan x\,\sec x $$, apply the product rule:
$$
p'=\frac{d}{dx}[\tan x]\cdot \sec x + \tan x\cdot \frac{d}{dx}[\sec x]
$$
We know $$\frac{d}{dx}[\tan x]=\sec^2 x$$ and $$\frac{d}{dx}[\sec x]=\sec x\tan x$$, so:
$$
p'=\sec^2 x\sec x+\tan x\cdot \sec x\tan x=\sec^3 x+\sec x\tan^2 x
$$
This can be factored to:
$$
p'=\sec x\left(\sec^2 x+\tan^2 x\right)
$$
(a) $$ h'(t) = \tan t + t \sec^2 t $$ (b) $$ g'(x) = 3x^2 \cos x - x^3 \sin x $$ (c) $$ v'(\theta) = 2(\theta+2)\sin \theta + (\theta+2)^2 \cos \theta $$ (d) $$ y' = \frac{(1+2x-x^2)\cos x - \sin x(2-2x)}{(1+2x-x^2)^2} $$ (e) $$ z' = 46(2\theta+3)^{22} \cos\bigl((2\theta+3)^{23}\bigr) $$ (f) $$ p' = \sec x (\sec^2 x + \tan^2 x) $$

2 Find the derivative of each function.
(a)
(b)
©
(d)
(e)
(f)

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