Using the product rule, where \( u = 5x^2 \) and \( v = e^{-x} \), we have:

["Using the Product Rule: Differentiating ( u = 5x^2 ) and ( v = e^{-x} )", "When tackling differentiation problems involving the product of two functions, the product rule is an essential tool in calculus. Today, we’ll explore how to apply the product rule to the functions ( u(x) = 5x^2 ) and ( v(x) = e^{-x} ), step by step. Mastering this rule not only strengthens your foundational calculus knowledge but also prepares you for more complex applications in physics, engineering, and beyond.", "---", "### What is the Product Rule?", "The product rule states that if you have two differentiable functions ( u(x) ) and ( v(x) ), the derivative of their product is:", "[\n(uv)' = u'v + uv'\n]", "That is, the derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.", "---", "### Applying the Product Rule to ( u = 5x^2 ) and ( v = e^{-x} )", "Let’s differentiate ( u ) and ( v ) individually first.", "- ( u = 5x^2 )\n Using the power rule,\n [\n u' = \frac{d}{dx}(5x^2) = 10x\n ]", "- ( v = e^{-x} )\n The exponential function with a linear argument differentiates to:\n [\n v' = \frac{d}{dx}(e^{-x}) = -e^{-x}\n ]", "---", "### Putting It All Together: The Product Rule in Action", "Now substitute into the product rule formula:", "[\n(uv)' = u'v + uv'\n]", "[\n\frac{d}{dx}(5x^2 \cdot e^{-x}) = (10x)(e^{-x}) + (5x^2)(-e^{-x})\n]", "[\n= 10x e^{-x} - 5x^2 e^{-x}\n]", "Factor out the common term ( e^{-x} ):", "[\n= e^{-x}(10x - 5x^2)\n]", "---", "### Final Answer", "[\n\frac{d}{dx}(5x^2 \cdot e^{-x}) = e^{-x}(10x - 5x^2)\n]", "---", "### Why This Matters", "Understanding the product rule and applying it correctly enables clear and precise differentiation of more complex functions common in scientific modeling, optimization, and dynamical systems. Using concrete examples like ( u = 5x^2 ) and ( v = e^{-x} ) grounds abstract calculus concepts in tangible results—ideal for both students and professionals alike.", "Mastering this technique unlocks greater fluency in calculus, empowering deeper insights and more confident problem-solving across disciplines.", "---", "Keywords: product rule, derivatives, calculus, differentiate, u = 5x², v = e⁻ˣ, u’ and v’, differentiation practice, math tutorial, exponential and polynomial functions, mathematical application", "Meta Description: Learn how to apply the product rule using ( u = 5x^2 ) and ( v = e^{-x} ). Step-by-step differentiation with clear explanation and final answer included. Perfect for students and math learners."]