Question

$\frac{d}{dx} [e^x(x^2 + 1)] =$

• A. $e^x(2x + x^2 + 1)$
• B. $e^x(2x - x^2 + 1)$
• C. $e^x(2x + x^3 + 1)$
• D. $e^{-x}(2x + x^2 + 1)$

Hint:

A useful shortcut for derivatives of the form $e^x \cdot f(x)$ is simply $e^x [f(x) + f'(x)]$. Here, $f(x) = x^2+1$ and $f'(x) = 2x$, giving the result directly.

Answer 1

The Correct Option is A

1. The Product Rule: If $y = u \cdot v$, then $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$. Let $u = e^x$ and $v = (x^2 + 1)$.

2. Differentiate Components: $\frac{du}{dx} = e^x$ $\frac{dv}{dx} = 2x\lt strong\gt 3. Combine using the Rule:\lt /strong\gt $ddx [e^x(x^2 + 1)] = e^x(2x) + (x^2 + 1)e^x

4. Factor out $e^x$: ddx [e^x(x^2 + 1)] = e^x(2x + x^2 + 1)$$ This matches Option (A).