Question

1. give the derivatives of the following functions. for this problem, we assume we know the derivatives of the basic six trigonometric functions. simplify where possible. a) f(x) = 3e^{2x} \tan x \frac{d(f(x))}{dx} = \underline{\hspace{5cm}}

Explanation:

Step1: Identify the product rule

The function \( f(x) = 3e^{2x}\tan x \) is a product of two functions, \( u = 3e^{2x} \) and \( v = \tan x \). The product rule states that \( (uv)' = u'v + uv' \).

Step2: Differentiate \( u = 3e^{2x} \)

Using the chain rule, the derivative of \( e^{ax} \) is \( ae^{ax} \). For \( u = 3e^{2x} \), \( u' = 3 \cdot 2e^{2x} = 6e^{2x} \).

Step3: Differentiate \( v = \tan x \)

The derivative of \( \tan x \) is \( \sec^2 x \), so \( v' = \sec^2 x \).

Step4: Apply the product rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule:
\[
f'(x) = u'v + uv' = 6e^{2x} \cdot \tan x + 3e^{2x} \cdot \sec^2 x
\]

Step5: Factor out common terms

Factor out \( 3e^{2x} \) from both terms:
\[
f'(x) = 3e^{2x}(2\tan x + \sec^2 x)
\]

Answer:

\( 3e^{2x}(2\tan x + \sec^2 x) \) (or equivalently \( 6e^{2x}\tan x + 3e^{2x}\sec^2 x \))