Question Statement

Find the first and second derivatives of the function:

$f(\theta )=\frac{1}{3+2\cos \theta }$

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Background and Explanation

This problem requires applying the chain rule to find the first derivative and the quotient rule (along with chain rule) for the second derivative. You should be familiar with derivatives of trigonometric functions: $\frac{d}{d\theta }(\cos \theta )=-\sin \theta$ and $\frac{d}{d\theta }(\sin \theta )=\cos \theta$.

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Solution

First Derivative

Begin by rewriting the function using negative exponents to apply the chain rule more easily:

$f(\theta )=(3+2\cos \theta {)}^{-1}$

Differentiate with respect to $\theta$ using the chain rule. The outer function is $u^{-1}$ and the inner function is $u=3+2\cos \theta$:

$f^{\prime }(\theta )=(-1)(3+2\cos \theta {)}^{-1-1}\cdot \frac{d}{d\theta }(3+2\cos \theta )$

Calculate the derivative of the inner function:

$\frac{d}{d\theta }(3+2\cos \theta )=0+2(-\sin \theta )=-2\sin \theta$

Substitute this back into the expression:

$f^{\prime }(\theta )=(-1)(3+2\cos \theta {)}^{-2}\cdot (-2\sin \theta )$

Simplify the signs ($(-1)\cdot (-2\sin \theta )=2\sin \theta$):

$f^{\prime }(\theta )=\frac{2\sin \theta }{(3+2\cos \theta {)}^2}$

Second Derivative

To find $f^{\prime \prime }(\theta )$, differentiate $f^{\prime }(\theta )$ using the quotient rule. Let:

• $u=2\sin \theta$ (numerator)
• $v=(3+2\cos \theta {)}^2$ (denominator)

Then:

• $u^{\prime }=2\cos \theta$
• $v^{\prime }=2(3+2\cos \theta )\cdot (-2\sin \theta )=-4\sin \theta (3+2\cos \theta )$

Apply the quotient rule formula $\frac{d}{d\theta }\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$:

$f^{\prime \prime }(\theta )=\frac{(3+2\cos \theta {)}^2\cdot (2\cos \theta )-2\sin \theta \cdot [-4\sin \theta (3+2\cos \theta )]}{(3+2\cos \theta {)}^4}$

Simplify the numerator (note the double negative becomes positive):

$=\frac{(3+2\cos \theta {)}^2(2\cos \theta )+8{\sin }^2\theta (3+2\cos \theta )}{(3+2\cos \theta {)}^4}$

Factor out $2(3+2\cos \theta )$ from the numerator:

$=\frac{2(3+2\cos \theta )\left[(3+2\cos \theta )\cos \theta +4{\sin }^2\theta \right]}{(3+2\cos \theta {)}^4}$

Cancel one factor of $(3+2\cos \theta )$ from numerator and denominator:

$f^{\prime \prime }(\theta )=\frac{2\left[(3+2\cos \theta )\cos \theta +4{\sin }^2\theta \right]}{(3+2\cos \theta {)}^3}$

Or equivalently:

$f^{\prime \prime }(\theta )=\frac{2\cos \theta (3+2\cos \theta )+8{\sin }^2\theta }{(3+2\cos \theta {)}^3}$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Trigonometric Derivatives: $\frac{d}{d\theta }(\cos \theta )=-\sin \theta$ and $\frac{d}{d\theta }(\sin \theta )=\cos \theta$

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Summary of Steps

1. Rewrite $f(\theta )$ as $(3+2\cos \theta {)}^{-1}$ to prepare for chain rule application
2. First derivative: Apply chain rule — multiply by $-1$, reduce exponent by $1$, and multiply by derivative of inner function $(-2\sin \theta )$
3. Simplify signs to get $f^{\prime }(\theta )=\frac{2\sin \theta }{(3+2\cos \theta {)}^2}$
4. Second derivative: Apply quotient rule to $f^{\prime }(\theta )$, identifying $u=2\sin \theta$ and $v=(3+2\cos \theta {)}^2$
5. Differentiate components: $u^{\prime }=2\cos \theta$ and $v^{\prime }=-4\sin \theta (3+2\cos \theta )$
6. Substitute into quotient rule formula and simplify the resulting expression
7. Factor out common term $2(3+2\cos \theta )$ from numerator and cancel with denominator
8. Final result: $f^{\prime \prime }(\theta )=\frac{2(3+2\cos \theta )\cos \theta +8{\sin }^2\theta }{(3+2\cos \theta {)}^3}$