Question:

Find the derivative of the function

Step-by-Step Solution:

To find the derivative of the given function using the chain rule, we will break it down into simpler functions and apply the chain rule to each part.

Step 1: Identify the outermost function and the inner composite function.

In this case, the outermost function is the fraction and the inner composite function is the ratio of two functions - the numerator and the denominator.

Step 2: Differentiate the numerator and denominator separately using the chain rule.

Let's begin by finding the derivative of the numerator, which is $\sin^2(x^2-3x)$.

Using the chain rule, we can differentiate the square of sine function with respect to its inner function.

$\frac{d}{dx}(\sin^2(u)) = 2\sin(u)\cos(u) \cdot \frac{du}{dx}$

Here, $u = x^2-3x$.

So, $\frac{d}{dx}\left(\sin^2(x^2-3x)\right) = 2\sin(x^2-3x)\cos(x^2-3x) \cdot \frac{d}{dx}(x^2-3x)$

Expanding the derivative of the denominator using the chain rule, we get:

$\frac{d}{dx}(x^2-3x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) = 2x - 3$

Substituting the above values, the derivative of the numerator becomes:

$\frac{d}{dx}\left(\sin^2(x^2-3x)\right) = 2\sin(x^2-3x)\cos(x^2-3x) \cdot (2x - 3)$

Step 3: Differentiate the denominator.

The denominator is $e^{2x^2}$.

To differentiate $e^u$, we multiply it by the derivative of its exponent.

$\frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx}$

In this case, $u = 2x^2$.

So, $\frac{d}{dx}(e^{2x^2}) = e^{2x^2} \cdot \frac{d}{dx}(2x^2)$

Again, using the chain rule, the derivative of the exponent becomes:

$\frac{d}{dx}(2x^2) = 2 \cdot \frac{d}{dx}(x^2) = 4x$

Substituting the above values, the derivative of the denominator becomes:

$\frac{d}{dx}(e^{2x^2}) = e^{2x^2} \cdot 4x = 4x \cdot e^{2x^2}$

Step 4: Apply the chain rule to find the derivative of the entire function.

Using the quotient rule, the derivative of the function $f(x) = \frac{\sin^2(x^2-3x)}{e^{2x^2}}$ can be written as:

Where $g(x) = \sin^2(x^2-3x)$ and $h(x) = e^{2x^2}$.

Substituting the expressions for $g'(x)$, $g(x)$, $h'(x)$, and $h(x)$, we get:

Simplifying the equation further, we have:

Finally, we can simplify the expression further, if desired, by factoring out the common terms $(2x - 3) \cdot e^{2x^2}$ in both the numerator:

Therefore, the derivative of the given function is: