f'(x) = rac{(x - 1)(2) - (2x + 3)(1)}{(x - 1)^2} = rac{2x - 2 - 2x - 3}{(x - 1)^2} = rac{-5}{(x - 1)^2}

["Understanding the Derivative: A Comprehensive Guide to f'(x) = −5 / (x − 1)²", "Calculating derivatives is a fundamental skill in calculus, and today we dive deep into a crucial expression:\n[\nf'(x) = \frac{(x - 1)(2) - (2x + 3)(1)}{(x - 1)^2} = \frac{2x - 2 - 2x - 3}{(x - 1)^2} = \frac{-5}{(x - 1)^2}\n]", "This simplified derivative, ( f'(x) = -\frac{5}{(x - 1)^2} ), plays an essential role in understanding function behavior—especially in fields like physics, economics, and optimization. In this article, we’ll explore how this derivative is derived, its meaning, how to use it, and why mastering this concept matters for advanced math and real-world applications.", "---", "### Step-by-Step Derivation of f'(x)", "Let’s first break down the process of getting to\n[\nf'(x) = \frac{-5}{(x - 1)^2}\n]\nfrom the quotient rule:\n[\nf'(x) = \frac{(x - 1)(2) - (2x + 3)(1)}{(x - 1)^2}\n]", "1. Apply the Quotient Rule:\n The derivative of ( \frac{u}{v} ) is ( \frac{u'v - uv'}{v^2} ).\n Here, ( u = 2(x - 1) ), ( v = (x - 1)^2 ).\n So ( u' = 2 ), ( v' = 2(x - 1) ).", "2. Plug in values:\n [\n f'(x) = \frac{[2 \cdot (x - 1)] - [(2x + 3) \cdot 2(x - 1)]}{(x - 1)^4}\n ]", "3. Expand the numerator:\n First term: ( 2(x - 1) = 2x - 2 )\n Second term:\n [\n (2x + 3) \cdot 2(x - 1) = 2(2x + 3)(x - 1)\n ]\n Expand:\n [\n 2[(2x)(x) + (2x)(-1) + (3)(x) + (3)(-1)] = 2(2x^2 - 2x + 3x - 3) = 2(2x^2 + x - 3) = 4x^2 + 2x - 6\n ]\n So numerator becomes:\n [\n (2x - 2) - (4x^2 + 2x - 6) = 2x - 2 - 4x^2 - 2x + 6 = -4x^2 + 4\n ]", "4. Simplify denominator:\n The denominator is ( (x - 1)^2 ), but in the quotient rule we square the denominator fully:\n [\n [(x - 1)^2]^2 = (x - 1)^4\n ]\n But since numerator is (-4x^2 + 4 = -4(x^2 - 1) = -4(x - 1)(x + 1)), we factor:", "5. Final simplification:\n [\n f'(x) = \frac{-4(x - 1)(x + 1)}{(x - 1)^4} = \frac{-4(x + 1)}{(x - 1)^3}\n ]", "Wait—this seems mismatched with the given simplified derivative. Let’s verify the earlier expression again.", "From the user-provided simplification:\n[\n\frac{(x - 1)(2) - (2x + 3)(1)}{(x - 1)^2} = \frac{2x - 2 - 2x - 3}{(x - 1)^2} = \frac{-5}{(x - 1)^2}\n]\nThis shortcut implies a numerator factored naively as ( -5 )—but algebraically, the numerator simplifies to ( -4x^2 + 4 ), not a constant. There is an inconsistency unless the numerator evaluates to (-5) identically.", "Let’s check:\nIs ( -4x^2 + 4 = -5 ) for all ( x )?\n[\n-4x^2 + 4 = -5 \Rightarrow -4x^2 = -9 \Rightarrow x^2 = \frac{9}{4} \Rightarrow x = \pm 1.5\n]\nOnly true at specific points, not generally.", "So the expression\n[\nf'(x) = \frac{-5}{(x - 1)^2}\n]\nis a simplified form valid only if the numerator evaluates to (-5) for all ( x <br/>\neq 1 )—which is not true.", "However, if this simplified derivative is manually claimed as accurate, it must be derived differently—perhaps assuming ( (x - 1)(2) - (2x + 3) = -5 ), ignoring higher-degree terms.", "Let’s recompute the numerator carefully:\n[\n2(x - 1) - (2x + 3) = 2x - 2 - 2x - 3 = -5\n]\nAh! This is key: the linear terms cancel, leaving a constant −5, but only because the variable terms cancel:\n- ( 2(x - 1) = 2x - 2 )\n- Subtract ( (2x + 3) ):\n [\n (2x - 2) - (2x + 3) = 2x - 2 - 2x - 3 = -5\n ]\nSo regardless of ( x ), the numerator simplifies algebraically to -5—but this is only true algebraically when treating it as a polynomial identity.", "But in standard calculus, the derivative is:\n[\nf'(x) = \frac{-4(x + 1)}{(x - 1)^3}\n]", "This suggests a possibility of typo in the original problem. However, if the given ( f'(x) = \frac{-5}{(x - 1)^2} ) is correct (possibly from a context where higher-degree terms cancel under specific conditions), then we proceed accordingly—but caution is needed.", "---", "### Why Simplification Works in Many Applications", "Even if the derivation seems messy or misleading, the simplified form ( f'(x) = -\frac{5}{(x - 1)^2} ) reveals powerful insights:", "- Domain: Undefined at ( x = 1 ), a vertical asymptote.\n- Sign Analysis: Since ( (x - 1)^2 > 0 ) for all ( x <br/>\ne 1 ), and numerator is negative, derivative is always negative—meaning ( f(x) ) is strictly decreasing near ( x = 1 ).\n- Behavior Near x = 1: As ( x \ o 1 ), ( f'(x) \ o -\infty ), indicating a vertical tangent or sharp corner.", "---", "### Who Uses This Derivative? Real-World Applications", "While such a derivative may seem abstract, derivatives of this form emerge in:\n- Optimization problems with rational cost or revenue functions.\n- Physics for defining inverse-square laws in modified coordinate systems.\n- Economics to model diminishing marginal returns with shifted inputs.", "Even if the algebra involves simplification beyond strict rules, recognizing patterns like constant numerators helps build intuition.", "---", "### Final Notes for Students and Professionals", "- Always verify algebra before accepting simplified forms.\n- Use tools like symbolic calculators (e.g., Wolfram Alpha) to confirm derivations.\n- Understand that simplified expressions often arise from domain restrictions or algebraic identity assumptions.", "---", "### Conclusion", "Though the path to\n[\nf'(x) = \frac{-5}{(x - 1)^2}\n]\nbegs closer inspection due to constant numerator simplification, this example highlights the importance of careful algebraic manipulation and awareness of domain behavior. Mastery of derivatives means not only computing them but recognizing where simplification is valid—and where deeper analysis is required.", "For learners, the key takeaway:\nAlgebra and calculus work hand-in-hand—never skip the steps, but always question the “why.”", "---", "Keywords: f'(x) = -5 / (x - 1)^2, derivative simplification, rational functions, calculus practice, vertical asymptote, application of quotient rule, function behavior, limit analysis, optimization derivative", "Meta Description:\nExplore the derivative ( f'(x) = \frac{-5}{(x - 1)^2} ) step-by-step, understand its derivation, significance, and real-world uses in calculus—while learning to verify simplifications and build strong mathematical intuition."]