Solved on Feb 03, 2024

Differentiate the function $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ .

STEP 1

Assumptions
1. The function to differentiate is $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ .
2. We will use the quotient rule for differentiation, which states that if $u(\theta)$ and $v(\theta)$ are differentiable functions of $\theta$ , then $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$ .

STEP 2

Identify the numerator $u(\theta)$ and the denominator $v(\theta)$ of the function.
$u(\theta) = \sec \theta$ $v(\theta) = 5 + \sec \theta$

STEP 3

Differentiate the numerator $u(\theta)$ with respect to $\theta$ .
$u'(\theta) = \frac{d}{d\theta}(\sec \theta)$

STEP 4

Use the derivative of $\sec \theta$ , which is $\sec \theta \tan \theta$ .
$u'(\theta) = \sec \theta \tan \theta$

STEP 5

Differentiate the denominator $v(\theta)$ with respect to $\theta$ .
$v'(\theta) = \frac{d}{d\theta}(5 + \sec \theta)$

STEP 6

Since the derivative of a constant is zero and using the derivative of $\sec \theta$ from STEP_4, we get:
$v'(\theta) = 0 + \sec \theta \tan \theta$

STEP 7

Now apply the quotient rule for differentiation.
$f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{v(\theta)^2}$

STEP 8

Substitute $u(\theta)$ , $u'(\theta)$ , $v(\theta)$ , and $v'(\theta)$ into the quotient rule formula.
$f'(\theta) = \frac{(\sec \theta \tan \theta)(5 + \sec \theta) - (\sec \theta)(\sec \theta \tan \theta)}{(5 + \sec \theta)^2}$

STEP 9

Expand the numerator of the derivative.
$f'(\theta) = \frac{5\sec \theta \tan \theta + \sec^2 \theta \tan \theta - \sec^2 \theta \tan \theta}{(5 + \sec \theta)^2}$

STEP 10

Notice that $\sec^2 \theta \tan \theta$ appears with opposite signs and will cancel each other out.
$f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$

STEP 11

This is the simplified form of the derivative.
$f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$
The derivative of the function $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ with respect to $\theta$ is $\frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$ .

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Differentiate the function $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ .

STEP 1

STEP 2

Identify the numerator $u(\theta)$ and the denominator $v(\theta)$ of the function.
$u(\theta) = \sec \theta$ $v(\theta) = 5 + \sec \theta$

STEP 3

Differentiate the numerator $u(\theta)$ with respect to $\theta$ .
$u'(\theta) = \frac{d}{d\theta}(\sec \theta)$

STEP 4

Use the derivative of $\sec \theta$ , which is $\sec \theta \tan \theta$ .
$u'(\theta) = \sec \theta \tan \theta$

STEP 5

Differentiate the denominator $v(\theta)$ with respect to $\theta$ .
$v'(\theta) = \frac{d}{d\theta}(5 + \sec \theta)$

STEP 6

Since the derivative of a constant is zero and using the derivative of $\sec \theta$ from STEP_4, we get:
$v'(\theta) = 0 + \sec \theta \tan \theta$

STEP 7

Now apply the quotient rule for differentiation.
$f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{v(\theta)^2}$

STEP 8

Substitute $u(\theta)$ , $u'(\theta)$ , $v(\theta)$ , and $v'(\theta)$ into the quotient rule formula.
$f'(\theta) = \frac{(\sec \theta \tan \theta)(5 + \sec \theta) - (\sec \theta)(\sec \theta \tan \theta)}{(5 + \sec \theta)^2}$

STEP 9

Expand the numerator of the derivative.
$f'(\theta) = \frac{5\sec \theta \tan \theta + \sec^2 \theta \tan \theta - \sec^2 \theta \tan \theta}{(5 + \sec \theta)^2}$

STEP 10

Notice that $\sec^2 \theta \tan \theta$ appears with opposite signs and will cancel each other out.
$f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$

STEP 11

This is the simplified form of the derivative.
$f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$
The derivative of the function $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ with respect to $\theta$ is $\frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$ .

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Differentiate the function f(θ)=sec⁡θ5+sec⁡θf(\theta) = \frac{\sec \theta}{5 + \sec \theta}f(θ)=5+secθsecθ​.

STEP 1

STEP 2

Identify the numerator u(θ) u(\theta) u(θ) and the denominator v(θ) v(\theta) v(θ) of the function.u(θ)=sec⁡θ u(\theta) = \sec \theta u(θ)=secθ v(θ)=5+sec⁡θ v(\theta) = 5 + \sec \theta v(θ)=5+secθ

STEP 3

Differentiate the numerator u(θ) u(\theta) u(θ) with respect to θ \theta θ.u′(θ)=ddθ(sec⁡θ) u'(\theta) = \frac{d}{d\theta}(\sec \theta) u′(θ)=dθd​(secθ)

STEP 4

Use the derivative of sec⁡θ \sec \theta secθ, which is sec⁡θtan⁡θ \sec \theta \tan \theta secθtanθ.u′(θ)=sec⁡θtan⁡θ u'(\theta) = \sec \theta \tan \theta u′(θ)=secθtanθ

STEP 5

Differentiate the denominator v(θ) v(\theta) v(θ) with respect to θ \theta θ.v′(θ)=ddθ(5+sec⁡θ) v'(\theta) = \frac{d}{d\theta}(5 + \sec \theta) v′(θ)=dθd​(5+secθ)

STEP 6

Since the derivative of a constant is zero and using the derivative of sec⁡θ \sec \theta secθ from STEP_4, we get:v′(θ)=0+sec⁡θtan⁡θ v'(\theta) = 0 + \sec \theta \tan \theta v′(θ)=0+secθtanθ

STEP 7

Now apply the quotient rule for differentiation.f′(θ)=u′(θ)v(θ)−u(θ)v′(θ)v(θ)2 f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{v(\theta)^2} f′(θ)=v(θ)2u′(θ)v(θ)−u(θ)v′(θ)​

STEP 8

STEP 9

Expand the numerator of the derivative.f′(θ)=5sec⁡θtan⁡θ+sec⁡2θtan⁡θ−sec⁡2θtan⁡θ(5+sec⁡θ)2 f'(\theta) = \frac{5\sec \theta \tan \theta + \sec^2 \theta \tan \theta - \sec^2 \theta \tan \theta}{(5 + \sec \theta)^2} f′(θ)=(5+secθ)25secθtanθ+sec2θtanθ−sec2θtanθ​

STEP 10

Notice that sec⁡2θtan⁡θ \sec^2 \theta \tan \theta sec2θtanθ appears with opposite signs and will cancel each other out.f′(θ)=5sec⁡θtan⁡θ(5+sec⁡θ)2 f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2} f′(θ)=(5+secθ)25secθtanθ​

STEP 11

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Differentiate the function $f(\theta) = \frac{\sec \theta}{5 + \sec \theta}$ .

Identify the numerator $u(\theta)$ and the denominator $v(\theta)$ of the function.
$u(\theta) = \sec \theta$ $v(\theta) = 5 + \sec \theta$

Differentiate the numerator $u(\theta)$ with respect to $\theta$ .
$u'(\theta) = \frac{d}{d\theta}(\sec \theta)$

Use the derivative of $\sec \theta$ , which is $\sec \theta \tan \theta$ .
$u'(\theta) = \sec \theta \tan \theta$

Differentiate the denominator $v(\theta)$ with respect to $\theta$ .
$v'(\theta) = \frac{d}{d\theta}(5 + \sec \theta)$

Since the derivative of a constant is zero and using the derivative of $\sec \theta$ from STEP_4, we get:
$v'(\theta) = 0 + \sec \theta \tan \theta$

Now apply the quotient rule for differentiation.
$f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{v(\theta)^2}$

Expand the numerator of the derivative.
$f'(\theta) = \frac{5\sec \theta \tan \theta + \sec^2 \theta \tan \theta - \sec^2 \theta \tan \theta}{(5 + \sec \theta)^2}$

Notice that $\sec^2 \theta \tan \theta$ appears with opposite signs and will cancel each other out.
$f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}$