Question

If the derivative of $(ax - 5)e^{3x}$ at $x = 0$ is $-13$, then the value of $a$ is equal to:

• A. $8$
• B. $-5$
• C. $5$
• D. $-2$
• E. $2$

Hint:

When evaluating derivatives at $x=0$, many terms simplify instantly (e.g., $e^0 = 1$, $ax = 0$). Instead of fully expanding the algebraic derivative, plug in zero as soon as possible to save time.

Answer 1

Answer: (E)

Concept: To find the derivative of a product of two functions, we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

Step 1: Differentiate the function.
Let $y = (ax - 5)e^{3x}$. Using the product rule where $u = ax - 5$ and $v = e^{3x}$: \[ \frac{dy}{dx} = \frac{d}{dx}(ax - 5) \cdot e^{3x} + (ax - 5) \cdot \frac{d}{dx}(e^{3x}) \] \[ \frac{dy}{dx} = (a)e^{3x} + (ax - 5)(3e^{3x}) \]

Step 2: Evaluate the derivative at $x = 0$.
Substitute $x = 0$ into the derivative expression: \[ \left[ \frac{dy}{dx} \right]_{x=0} = a \cdot e^0 + (a(0) - 5)(3 \cdot e^0) \] \[ -13 = a(1) + (-5)(3) \] \[ -13 = a - 15 \]

Step 3: Solve for $a$.
\[ a = -13 + 15 \] \[ a = 2 \]