C''(x) = rac{d}{dx}(e^{-x}(10x - 5x^2)) = e^{-x}(-10x + 5x^2) + e^{-x}(10 - 10x).
["Exactly Calculating the Derivative: C''(x) = \frac{d}{dx}\left(e^{-x}(10x - 5x^2)\right)", "When analyzing functions involving products of exponential and polynomial expressions, mastering differentiation techniques is essential. One powerful example is computing the second derivative of ( C(x) = e^{-x}(10x - 5x^2) ). This expression commonly appears in physics, economics, and calculus problems—especially when modeling decaying processes with quadratic trends.", "This article breaks down the full derivation of ( C''(x) ), including careful application of the product rule, simplification, and verification of each step. We’ll explore:", "- The original function and the significance of the product of exponential decay with a quadratic term\n- Application of the product rule for differentiation\n- Algebraic simplification steps leading to the final expression\n- Interpretation of the result and practical relevance", "---", "### Understanding C(x)", "Let’s begin with the function:", "[\nC(x) = e^{-x}(10x - 5x^2)\n]", "This represents a quantity diminishing over time due to the exponential factor ( e^{-x} ), modulated by a quadratic waveform ( 10x - 5x^2 ). Such forms often model resonant decay, profit curves, or transient signals where growth peaks before decay dominates.", "---", "### First Derivative: ( C'(x) )", "To compute the second derivative, we first need the first derivative. Because ( C(x) ) is a product of two functions:", "- ( u(x) = e^{-x} )\n- ( v(x) = 10x - 5x^2 )", "We apply the product rule:", "[\nC'(x) = u'(x)v(x) + u(x)v'(x)\n]", "First, compute derivatives:", "- ( u'(x) = -e^{-x} )\n- ( v'(x) = 10 - 10x )", "Now substitute:", "[\nC'(x) = (-e^{-x})(10x - 5x^2) + e^{-x}(10 - 10x)\n]", "Factor out ( e^{-x} ):", "[\nC'(x) = e^{-x} \left[ -(10x - 5x^2) + (10 - 10x) \right]\n]", "Simplify inside the brackets:", "[\n-(10x - 5x^2) + (10 - 10x) = -10x + 5x^2 + 10 - 10x = 5x^2 - 20x + 10\n]", "So,", "[\nC'(x) = e^{-x}(5x^2 - 20x + 10)\n]", "---", "### Second Derivative: ( C''(x) )", "Now differentiate ( C'(x) = e^{-x}(5x^2 - 20x + 10) ) using the same product rule.", "Let:", "- ( u(x) = e^{-x} \Rightarrow u'(x) = -e^{-x} )\n- ( v(x) = 5x^2 - 20x + 10 \Rightarrow v'(x) = 10x - 20 )", "Apply the product rule:", "[\nC''(x) = u'(x)v(x) + u(x)v'(x) = (-e^{-x})(5x^2 - 20x + 10) + e^{-x}(10x - 20)\n]", "Factor out ( e^{-x} ):", "[\nC''(x) = e^{-x} \left[ -(5x^2 - 20x + 10) + (10x - 20) \right]\n]", "Expand inside the brackets:", "[\n-(5x^2 - 20x + 10) + (10x - 20) = -5x^2 + 20x - 10 + 10x - 20 = -5x^2 + 30x - 30\n]", "Therefore, the final expression is:", "[\nC''(x) = e^{-x}(-5x^2 + 30x - 30)\n]", "---", "### Rewriting the Result", "We can factor the quadratic:", "[\nC''(x) = e^{-x} \cdot (-5)(x^2 - 6x + 6) = -5e^{-x}(x^2 - 6x + 6)\n]", "Thus,", "[\n\boxed{C''(x) = e^{-x}(-5x^2 + 30x - 30)}\n]", "---", "### Interpretation and Applications", "The second derivative provides insight into the concavity of ( C(x) ):", "- A positive ( C''(x) ) indicates concave up (increasing slope)\n- A negative ( C''(x) ) indicates concave down (decreasing slope)", "Finding ( C''(x) ) helps identify local maxima, minima, and inflection points by solving ( C''(x) = 0 ):", "[\n-5x^2 + 30x - 30 = 0 \Rightarrow x^2 - 6x + 6 = 0\n]", "Solving gives:", "[\nx = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3}\n]", "These critical points define where the rate of decrease (or increase) toggles due to concavity changes—valuable in optimization and modeling.", "---", "### Final Notes", "This problem showcases:", "- The power of the product rule in handling composite expressions\n- Strategic factoring to simplify results\n- The utility of second derivatives in analyzing function behavior", "Whether you're solving differential equations, modeling physical decay, or optimizing quadratic-exponential systems, understanding such derivatives equips you with a robust analytical tool.", "---", "Keywords:\n( \frac{d}{dx}(e^{-x}(10x - 5x^2)) ), derivative calculation, product rule, second derivative, ( C''(x) ), calculus tutorial, exponential decay, polynomial derivative, concavity, optimization, ( C''(x) = e^{-x}(-5x^2 + 30x - 30) )"]
