Problem 12 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\left(9 x^{2}-6 x+2\right) e^{3 x}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = (27x^2 + 12x)e^{3x} \).

Step 1: Identify the Function Type

The given function is a product of two functions: a polynomial \(9x^2 - 6x + 2\) and an exponential function \(e^{3x}\). Therefore, we should use the product rule to find the derivative.

Step 2: Recall the Product Rule

The product rule states that if \(y = u(x) \, v(x)\), then \(\frac{dy}{dx} = u'(x) \, v(x) + u(x) \, v'(x)\). Here, let \(u(x) = 9x^2 - 6x + 2\) and \(v(x) = e^{3x}\).

Step 3: Differentiate the First Function

Find the derivative of the polynomial. \[u(x) = 9x^2 - 6x + 2 \] Differentiate to get \[ u'(x) = 18x - 6 \]

Step 4: Differentiate the Second Function

Find the derivative of the exponential function. \[v(x) = e^{3x} \] Differentiating gives us \[ v'(x) = 3e^{3x} \] using chain rule.

Step 5: Apply the Product Rule

Plug \(u(x), u'(x), v(x), \) and \(v'(x)\) into the product rule formula:\[ \frac{dy}{dx} = (18x - 6)e^{3x} + (9x^2 - 6x + 2)(3e^{3x}) \]

Step 6: Simplify the Expression

Combine and simplify the terms:\[ \frac{dy}{dx} = (18x - 6)e^{3x} + (27x^2 - 18x + 6)e^{3x} \]\[ = (27x^2 + 0x)e^{3x} \]\[ = 27x^2 e^{3x} \]

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus used to find the derivative of a product of two differentiable functions. If we have two functions, say \( u(x) \) and \( v(x) \), their product is denoted as \( y = u(x) \cdot v(x) \). According to the product rule, the derivative of \( y \) with respect to \( x \) is:

• \( \frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

This captures how both functions are changing, providing us with a complete view of the derivative. By applying the product rule, you can effectively deal with derivatives involving multiple functions interacting through multiplication.
In our exercise, the polynomial function \( 9x^2 - 6x + 2 \) and the exponential function \( e^{3x} \) constitute \( u(x) \) and \( v(x) \), respectively. Using the product rule allows us to successfully determine the derivative of the entire expression.

Polynomial Functions

Polynomial functions are expressions consisting of variables raised to whole-number exponents combined with coefficients, and they take the general form, \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 \). These functions are considered very smooth and have well-known properties such as continuity, differentiability, and familiar derivative rules.

• The derivative of a polynomial function can be found using the power rule, which states that for \( ax^n \), \( \frac{d}{dx}(ax^n) = n\cdot ax^{n-1} \).

In our exercise, the polynomial function \( 9x^2 - 6x + 2 \) is present. Differentiating it using the power rule, term by term, we get:

• \( \frac{d}{dx}(9x^2) = 18x \)
• \( \frac{d}{dx}(-6x) = -6 \)

Combining these results gives us the derivative: \( 18x - 6 \).

Exponential Functions

Exponential functions have the form \( f(x) = a^{g(x)} \), where \( e^x \) is the most common base. These functions exhibit rapid growth or decay and are vital in modeling various real-world situations. The derivative of \( e^{g(x)} \) is specific due to its unique property, where the derivative is the function itself multiplied by the derivative of the exponent.

• If \( f(x) = e^{g(x)} \), then \( f'(x) = g'(x) \cdot e^{g(x)} \)

This is particularly useful in calculus due to its simplicity.
For the function \( e^{3x} \) in our problem, the chain rule aids in finding the derivative. We see:

• \( \frac{d}{dx}(e^{3x}) = 3e^{3x} \)

where the \( 3 \) comes from the derivative of \( 3x \). This highlights how differentiation of exponential functions can rely on exponential rules combined with the chain rule.

Chain Rule

The chain rule is crucial for differentiating composite functions, where one function is nested inside another. The chain rule states that if a function \( y = f(g(x)) \) is composed, then the derivative \( y' \) is derived as follows:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

This essentially unravels the composition by finding the derivative of the outer function at the inner function times the derivative of the inner function.
In the function \( e^{3x} \) observed in our problem, \( g(x) = 3x \), and \( f(u) = e^u \) where \( u = 3x \). By differentiating, we first find \( g'(x) = 3 \) and then \( f'(u) = e^u \) leading to:

• \( \frac{d}{dx}(e^{3x}) = 3 \cdot e^{3x} \)

The chain rule seamlessly integrates these layers, enabling us to take derivatives of more complex expressions where functions are interdependent.