\( f'(2) = 3(2)^2 - 6(2) + 5 = 12 - 12 + 5 = 5 \)

["# Solving ( f'(2) = 3(2)^2 - 6(2) + 5 = 5 ): A Step-by-Step Guide to Finding the Derivative", "Understanding how to compute derivatives and evaluate them at specific points is a fundamental skill in calculus. In this article, we explore how to find ( f'(2) ) using the derivative formula provided:\n[\nf'(2) = 3(2)^2 - 6(2) + 5\n]\nWe’ll break down the calculation, explain the significance of this result in calculus, and clarify why this form appears in many algebra and derivatives practice problems.", "---", "## Introduction to Derivatives and ( f'(x) )", "In calculus, the derivative ( f'(x) ) represents the instantaneous rate of change of the function ( f(x) ) at any point ( x ). Computing ( f'(x) ) involves applying differentiation rules to find an algebraic expression, then substituting the desired value of ( x )—in this case, ( x = 2 ).", "The expression ( f'(2) = 3(2)^2 - 6(2) + 5 = 12 - 12 + 5 = 5 ) shows a specific evaluation: after differentiating a function with a given polynomial expression, we confirm that the slope of the tangent line at ( x = 2 ) is 5.", "---", "## Step-by-Step Breakdown of ( f'(2) = 3(2)^2 - 6(2) + 5 )", "### Step 1: Identify the derivative expression\nThe problem presents the derivative at ( x = 2 ) as:\n[\nf'(2) = 3(2)^2 - 6(2) + 5\n]\nThis expression stems from differentiating a general quadratic form. Assume the original function is:\n[\nf(x) = 3x^3 - 6x^2 + 5x + C\n]\nUsing standard differentiation rules:\n- The derivative of ( 3x^3 ) is ( 3 \cdot 3x^{3-1} = 9x^2 )\n- The derivative of ( -6x^2 ) is ( -6 \cdot 2x = -12x )\n- The derivative of ( 5x ) is ( 5 )\n- The derivative of constant ( C ) is 0", "So:\n[\nf'(x) = 9x^2 - 12x + 5\n]", "### Step 2: Plug in ( x = 2 )\nNow substitute ( x = 2 ) into the derivative:\n[\nf'(2) = 9(2)^2 - 12(2) + 5\n]\nCalculate term by term:\n- ( 9(2)^2 = 9 \cdot 4 = 36 )\n- ( -12(2) = -24 )\n- Constant: ( +5 )", "Add:\n[\nf'(2) = 36 - 24 + 5 = 17\n]", "Wait — this gives 17, not 5! But the original claim is ( f'(2) = 5 ), meaning our example derivative differs from standard forms.", "### Step 3: Reconciling the Expression", "The expression ( f'(2) = 3(2)^2 - 6(2) + 5 = 12 - 12 + 5 = 5 ) must correspond to a specific differentiation setup, perhaps from a simplified model or typo in coefficient placement.", "Let’s reverse-engineer: for ( f'(2) = 5 ), suppose ( f'(x) = ax^2 + bx + c ), and:\n[\nf'(2) = a(4) + b(2) + c = 5\n]\nTesting ( a = 3, b = -6, c = 5 ):\n[\n3(4) -6(2) + 5 = 12 -12 + 5 = 5\n]\nThus, the derivative function ( f'(x) = 3x^2 - 6x + 5 ) produces ( f'(2) = 5 ).", "---", "## Why This Form Appears in Practice", "Expressions like this commonly appear when:\n- The function is a cubic polynomial ( f(x) = ax^3 + bx^2 + cx + d ), whose derivative is a quadratic.\n- Coefficients reflect a simplified or example function used in learning materials.\n- Derivatives are evaluated at small integers like ( x = 2 ), making arithmetic quick and error-minimizing.", "---", "## Summary: Evaluating ( f'(2) ) from the Given Formula", "We verify:\n[\nf'(2) = 3(2)^2 - 6(2) + 5 = 12 - 12 + 5 = 5\n]\nThis demonstrates:\n- Substitution into a derivative polynomial.\n- Stepwise evaluation using arithmetic and exponentiation.\n- How values at specific points capture important function behavior like slope.", "---", "## Final Thoughts on Computing Derivatives at Specific Points", "Calculating ( f'(x) ) and evaluating at ( x = a ) helps visualize how functions change locally. When dealing with algebraic expressions, double-check signs, exponents, and coefficients to avoid common errors. Mastering this step ensures accuracy in solving optimization, curve sketching, and real-world modeling problems.", "If you're practicing derivatives, try:\n- Differentiate general forms, then compute ( f'(a) ) manually.\n- Use synthetic differentiation rules for efficiency.\n- Confirm results with numerical approximations or graphing tools.", "---", "Keywords: ( f'(2) ), derivative calculation, evaluate derivative, calculate ( f'(2) ), how to compute derivatives, slope at point, algebraic differentiation, calculus practice, derivative examples, instantaneous rate of change.", "---", "Related Reading:\n- How to Find the Derivative of Any Polynomial\n- Understanding the Mean Value Theorem and Derivatives\n- Tips for Handling Derivative Practice Problems", "---", "Note: The correct evaluation of ( f'(2) = 3(2)^2 - 6(2) + 5 ) yields 5, verifying this exact expression represents a valid derivative configuration. Always verify coefficient placement and operations when solving derivative problems."]