We are tasked with differentiating the function $y = e^{9x} \sin x$ with respect to $x$.

Solution:

This involves the product rule since $y$ is a product of $e^{9x}$ and $\sin x$.

The product rule states: $\frac{d}{dx}(u \cdot v) = u'v + uv'$ Here:

• $u = e^{9x}$, so $u' = \frac{d}{dx}(e^{9x}) = 9e^{9x}$ (using the chain rule).
• $v = \sin x$, so $v' = \frac{d}{dx}(\sin x) = \cos x$.

Now applying the product rule: $\frac{dy}{dx} = u'v + uv' = (9e^{9x})(\sin x) + (e^{9x})(\cos x).$

Final Answer:

$\frac{dy}{dx} = 9e^{9x} \sin x + e^{9x} \cos x.$

Would you like further details or a breakdown of the differentiation rules?

Related Questions:

1. How does the chain rule apply in differentiating exponential functions like $e^{9x}$?
2. What is the significance of the product rule in calculus?
3. Can you differentiate $e^{ax} \cos x$ using a similar approach?
4. How does one simplify expressions involving derivatives of trigonometric and exponential functions?
5. What are other scenarios where the product rule and chain rule must be combined?

Tip:

Always identify whether a function is a product, quotient, or composition before applying differentiation rules to avoid errors.