To find [tex]\(\frac{dy}{dx}\)[/tex] for [tex]\(y = x^3 \tan(2x)\)[/tex], we must use both the product rule and the chain rule in differentiation. Here’s a detailed step-by-step solution:
1. Identify the functions:
- Let's set [tex]\(u = x^3\)[/tex]
- Let [tex]\(v = \tan(2x)\)[/tex]
So [tex]\(y = u \cdot v\)[/tex].
2. Apply the product rule:
- The product rule states that [tex]\(\frac{d}{dx} [u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\)[/tex].
3. Differentiate [tex]\(u = x^3\)[/tex]:
- [tex]\(\frac{du}{dx} = 3x^2\)[/tex]
4. Differentiate [tex]\(v = \tan(2x)\)[/tex] using the chain rule:
- Start by differentiating [tex]\(\tan(2x)\)[/tex] with respect to [tex]\(2x\)[/tex], giving [tex]\(\sec^2(2x)\)[/tex].
- Then differentiate [tex]\(2x\)[/tex] with respect to [tex]\(x\)[/tex], giving [tex]\(2\)[/tex].
- Hence, [tex]\(\frac{d}{dx}[\tan(2x)] = 2 \sec^2(2x)\)[/tex].
5. Combine these results using the product rule:
- The derivative [tex]\(\frac{dy}{dx}\)[/tex] will be given by:
[tex]\[
\frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}
\][/tex]
- Substitute [tex]\(u\)[/tex], [tex]\(\frac{du}{dx}\)[/tex], [tex]\(v\)[/tex], and [tex]\(\frac{dv}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = x^3 \cdot 2\sec^2(2x) + \tan(2x) \cdot 3x^2
\][/tex]
6. Simplify:
- This becomes:
[tex]\[
\frac{dy}{dx} = 2x^3 \sec^2(2x) + 3x^2 \tan(2x)
\][/tex]
Comparing this with the given options, the correct answer is:
a) [tex]\(3 x^2 \tan 2 x + 2 x^3 \sec^2 2 x\)[/tex]
