Question:

Let $f(x) = \frac{\sin^3(x) + 2x}{e^{2x}\cos(x)}$. Find $f'(x)$ using the product and quotient rules.

Answer:

To find the derivative of $f(x) = \frac{\sin^3(x) + 2x}{e^{2x}\cos(x)}$ using the product and quotient rules, we will differentiate the numerator and denominator separately and then apply the rules accordingly.

Let's start with differentiating the numerator:

1. We have $\frac{d}{dx}\sin^3(x)$. Applying the chain rule, we consider $\sin^3(x)$ as $(\sin(x))^3$. The derivative of $\sin^3(x)$ is $3(\sin^2(x))\cos(x)$.

2. Next, we differentiate $2x$ with respect to $x$, which gives $2$.

Now, let's differentiate the denominator:

3. We have $\frac{d}{dx}(e^{2x}\cos(x))$. Applying the product rule, we differentiate $e^{2x}$ and $\cos(x)$ separately and keep the other term intact.

• Applying the chain rule, the derivative of $e^{2x}$ is $2e^{2x}$.
• Differentiating $\cos(x)$ gives $-\sin(x)$.

Now, we can use the product and quotient rules to find the derivative of $f(x)$:

4. Apply the quotient rule: $\frac{d}{dx} \left( \frac{\sin^3(x) + 2x}{e^{2x}\cos(x)} \right) = \frac{(3\sin^2(x)\cos(x))(e^{2x}\cos(x)) - (2)(\sin^3(x) + 2x)(2e^{2x} - \sin(x))}{(e^{2x}\cos(x))^2}$.

5. Simplify the expression: $\frac{3\sin^2(x)\cos^2(x)e^{2x} - 2(2e^{2x}\sin^3(x) - 2x(2e^{2x} - \sin(x)))}{(e^{2x}\cos(x))^2}$.

6. Further simplification: $\frac{3\sin^2(x)\cos^2(x)e^{2x} - 4e^{2x}\sin^3(x) + 4x(2e^{2x} - \sin(x))}{(e^{2x}\cos(x))^2}$.

Therefore, $f'(x) = \frac{3\sin^2(x)\cos^2(x)e^{2x} - 4e^{2x}\sin^3(x) + 4x(2e^{2x} - \sin(x))}{(e^{2x}\cos(x))^2}$.

Hence, we have found the derivative of $f(x)$ using the product and quotient rules.