Question Statement

Differentiate the function $f (θ) = (2 θ + 1)^{3} tan^{2} θ$ with respect to $θ$ .

Background and Explanation

This problem involves differentiating a product of two composite functions: a polynomial raised to a power $(2 θ + 1)^{3}$ and a squared trigonometric function $tan^{2} θ$ . You will need to apply the product rule in combination with the chain rule to handle each component.

Solution

We begin by recognizing that $f (θ)$ is a product of two functions:

$u = (2 θ + 1)^{3}$
$v = tan^{2} θ$

To find $f^{'} (θ)$ , we apply the product rule: $\frac{d}{d θ} (u \cdot v) = \frac{d u}{d θ} \cdot v + u \cdot \frac{d v}{d θ}$

$f^{'} (θ) = \frac{d}{d θ} (2 θ + 1)^{3} \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot \frac{d}{d θ} (tan^{2} θ) = 3 (2 θ + 1)^{2} \cdot \frac{d}{d θ} (2 θ + 1) \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot 2 tan^{2 - 1} θ \cdot \frac{d}{d θ} (tan θ) = 3 (2 θ + 1)^{2} \cdot (2 + 0) \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot 2 tan θ \cdot sec^{2} θ = 6 (2 θ + 1)^{2} tan^{2} θ + 2 (2 θ + 1)^{3} tan θ sec^{2} θ = 2 (2 θ + 1)^{2} tan θ [3 tan θ + (2 θ + 1) sec^{2} θ]$

Step-by-step reasoning:

Apply the product rule: We differentiate the first function $(2 θ + 1)^{3}$ while keeping the second $tan^{2} θ$ unchanged, then add the first function unchanged times the derivative of the second.
Differentiate $(2 θ + 1)^{3}$ : Using the chain rule, we bring down the power 3, reduce the power by 1, then multiply by the derivative of the inner function $(2 θ + 1)$ , which is 2.
Differentiate $tan^{2} θ$ : Using the chain rule on this composite function, we treat it as $(tan θ)^{2}$ . We bring down the power 2, reduce the power by 1 to get $tan θ$ , then multiply by the derivative of $tan θ$ , which is $sec^{2} θ$ .
Simplify: Multiply the constants $3 \times 2 = 6$ in the first term.
Factor: Extract the common factors $2 (2 θ + 1)^{2} tan θ$ from both terms to obtain the most compact form of the answer.

Key Formulas or Methods Used

Product Rule: $\frac{d}{d θ} [u (θ) \cdot v (θ)] = u^{'} (θ) \cdot v (θ) + u (θ) \cdot v^{'} (θ)$
Chain Rule: $\frac{d}{d θ} [f (g (θ))] = f^{'} (g (θ)) \cdot g^{'} (θ)$
Power Rule: $\frac{d}{d θ} [θ^{n}] = n θ^{n - 1}$
Derivative of tangent: $\frac{d}{d θ} [tan θ] = sec^{2} θ$

Summary of Steps

Identify the two functions being multiplied: $u = (2 θ + 1)^{3}$ and $v = tan^{2} θ$
Find $\frac{d u}{d θ} = 3 (2 θ + 1)^{2} \cdot 2 = 6 (2 θ + 1)^{2}$ using the chain rule
Find $\frac{d v}{d θ} = 2 tan θ \cdot sec^{2} θ$ using the chain rule
Apply the product rule: $f^{'} (θ) = \frac{d u}{d θ} \cdot v + u \cdot \frac{d v}{d θ}$
Substitute the derivatives and simplify the expression
Factor out $2 (2 θ + 1)^{2} tan θ$ to obtain the final factored form

Solution

We begin by recognizing that

f (θ)

is a product of two functions:

u = (2 θ + 1)^{3}

v = tan^{2} θ

To find

f^{'} (θ)

, we apply the product rule:

\frac{d}{d θ} (u \cdot v) = \frac{d u}{d θ} \cdot v + u \cdot \frac{d v}{d θ}

f^{'} (θ) = \frac{d}{d θ} (2 θ + 1)^{3} \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot \frac{d}{d θ} (tan^{2} θ) = 3 (2 θ + 1)^{2} \cdot \frac{d}{d θ} (2 θ + 1) \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot 2 tan^{2 - 1} θ \cdot \frac{d}{d θ} (tan θ) = 3 (2 θ + 1)^{2} \cdot (2 + 0) \cdot tan^{2} θ + (2 θ + 1)^{3} \cdot 2 tan θ \cdot sec^{2} θ = 6 (2 θ + 1)^{2} tan^{2} θ + 2 (2 θ + 1)^{3} tan θ sec^{2} θ = 2 (2 θ + 1)^{2} tan θ [3 tan θ + (2 θ + 1) sec^{2} θ]

Step-by-step reasoning:

Apply the product rule: We differentiate the first function $(2 θ + 1)^{3}$ while keeping the second $tan^{2} θ$ unchanged, then add the first function unchanged times the derivative of the second.

Differentiate $(2 θ + 1)^{3}$ : Using the chain rule, we bring down the power 3, reduce the power by 1, then multiply by the derivative of the inner function $(2 θ + 1)$ , which is 2.

Differentiate $tan^{2} θ$ : Using the chain rule on this composite function, we treat it as $(tan θ)^{2}$ . We bring down the power 2, reduce the power by 1 to get $tan θ$ , then multiply by the derivative of $tan θ$ , which is $sec^{2} θ$ .

Simplify: Multiply the constants $3 \times 2 = 6$ in the first term.

Factor: Extract the common factors $2 (2 θ + 1)^{2} tan θ$ from both terms to obtain the most compact form of the answer.

Summary of Steps

Identify the two functions being multiplied:

u = (2 θ + 1)^{3}

and

v = tan^{2} θ

Find

\frac{d u}{d θ} = 3 (2 θ + 1)^{2} \cdot 2 = 6 (2 θ + 1)^{2}

using the chain rule

Find

\frac{d v}{d θ} = 2 tan θ \cdot sec^{2} θ

using the chain rule

Apply the product rule:

f^{'} (θ) = \frac{d u}{d θ} \cdot v + u \cdot \frac{d v}{d θ}

Substitute the derivatives and simplify the expression

Factor out

2 (2 θ + 1)^{2} tan θ

to obtain the final factored form

2.7 Q-4

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.7 Q-4

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps