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

Studdy Solution

STEP 1

Assumptions
1. The function to differentiate is f(θ)=sec⁡θ5+sec⁡θ f(\theta) = \frac{\sec \theta}{5 + \sec \theta} .
2. We will use the quotient rule for differentiation, which states that if u(θ) u(\theta) and v(θ) v(\theta) are differentiable functions of θ \theta , then (uv)′=u′v−uv′v2 \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} .

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

Differentiate the denominator v(θ) v(\theta) with respect to θ \theta .
v′(θ)=ddθ(5+sec⁡θ) 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⁡θ \sec \theta from STEP_4, we get:
v′(θ)=0+sec⁡θtan⁡θ v'(\theta) = 0 + \sec \theta \tan \theta

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}

STEP 8

Substitute u(θ) u(\theta) , u′(θ) u'(\theta) , v(θ) v(\theta) , and v′(θ) v'(\theta) into the quotient rule formula.
f′(θ)=(sec⁡θtan⁡θ)(5+sec⁡θ)−(sec⁡θ)(sec⁡θtan⁡θ)(5+sec⁡θ)2 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′(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}

STEP 10

Notice that sec⁡2θtan⁡θ \sec^2 \theta \tan \theta 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}

STEP 11

This is the simplified form of the derivative.
f′(θ)=5sec⁡θtan⁡θ(5+sec⁡θ)2 f'(\theta) = \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2}
The derivative of the function f(θ)=sec⁡θ5+sec⁡θ f(\theta) = \frac{\sec \theta}{5 + \sec \theta} with respect to θ \theta is 5sec⁡θtan⁡θ(5+sec⁡θ)2 \frac{5\sec \theta \tan \theta}{(5 + \sec \theta)^2} .

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