Question

If $f(x)=(2x+3)e^{5x}$, then $f^{\prime\prime}(1)-10f^{\prime}(1)$ is equal to

• A. $250e^{5}$
• B. $125e^{5}$
• C. $25e^{5}$
• D. $-25e^{5}$
• E. $-125e^{5}$

Hint:

Logic Tip: Always factor out the exponential term $e^{kx}$ immediately after taking a derivative. Distributing it creates multiple product rule scenarios in the next step, massively increasing the risk of arithmetic errors.

Answer 1

Answer: (E)

Concept:
This problem requires successive differentiation using the Product Rule: $(uv)' = u'v + uv'$, alongside the Chain Rule for exponential functions: $\frac{d}{dx}(e^{kx}) = ke^{kx}$.
Step 1: Calculate the first derivative, $f'(x)$.
Let $u = 2x+3$ and $v = e^{5x}$. $u' = 2$ and $v' = 5e^{5x}$. Using the product rule: $$f'(x) = (2)(e^{5x}) + (2x+3)(5e^{5x})$$ Factor out $e^{5x}$ to simplify for the next derivative: $$f'(x) = e^{5x} [2 + 5(2x+3)]$$ $$f'(x) = e^{5x} [2 + 10x + 15]$$ $$f'(x) = (10x + 17)e^{5x}$$
Step 2: Calculate the second derivative, $f''(x)$.
Apply the product rule again to $f'(x)$. Let $u = 10x+17$ and $v = e^{5x}$. $u' = 10$ and $v' = 5e^{5x}$. $$f''(x) = (10)(e^{5x}) + (10x+17)(5e^{5x})$$ Factor out $e^{5x}$: $$f''(x) = e^{5x} [10 + 5(10x+17)]$$ $$f''(x) = e^{5x} [10 + 50x + 85]$$ $$f''(x) = (50x + 95)e^{5x}$$
Step 3: Construct the target algebraic expression.
We need the value of $f''(x) - 10f'(x)$. Let's calculate this generally before plugging in $x=1$: $$f''(x) - 10f'(x) = (50x + 95)e^{5x} - 10 \cdot [(10x + 17)e^{5x}]$$ Factor out $e^{5x}$: $$= e^{5x} [ (50x + 95) - (100x + 170) ]$$ $$= e^{5x} [ -50x - 75 ]$$
Step 4: Evaluate the expression at $x=1$.
Substitute $x = 1$ into the simplified expression: $$f''(1) - 10f'(1) = e^{5(1)} [ -50(1) - 75 ]$$ $$= e^5 [ -50 - 75 ]$$ $$= -125e^5$$