Derivative Rules Half Sheet Worksheet
d/dx
Calculus Derivative Drill
Front • Inverse Trig & Absolute Value Rules
Name:
Date:
Per:
1 Chain Rule with Inverse Trig
Show all differentiation steps
Find \(\frac{dy}{dx}\) for the function \(y = \arctan\left(2x^3 - 5x\right)\). Express your final derivative in simplest factored form.
Final Answer \(\frac{dy}{dx} =\):
2 Absolute Value of a Trigonometric Function
Differentiate & evaluate
Let \(f(x) = \left|\cos(2x)\right|\). Find an expression for \(f'(x)\) for all \(x\) where \(f\) is differentiable, then evaluate \(f'\left(\frac{\pi}{6}\right)\).
\(f'(x) =\):
\(f'\left(\frac{\pi}{6}\right) =\):
Cut along line for two half-sheet copies
d/dx
Calculus Derivative Drill
Front • Inverse Trig & Absolute Value Rules
Name:
Date:
Per:
1 Chain Rule with Inverse Trig
Show all differentiation steps
Find \(\frac{dy}{dx}\) for the function \(y = \arctan\left(2x^3 - 5x\right)\). Express your final derivative in simplest factored form.
Final Answer \(\frac{dy}{dx} =\):
2 Absolute Value of a Trigonometric Function
Differentiate & evaluate
Let \(f(x) = \left|\cos(2x)\right|\). Find an expression for \(f'(x)\) for all \(x\) where \(f\) is differentiable, then evaluate \(f'\left(\frac{\pi}{6}\right)\).
\(f'(x) =\):
\(f'\left(\frac{\pi}{6}\right) =\):
f⁻¹
Calculus Derivative Drill
Back • Derivatives of Inverse Functions
Rule: \((f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}\)
3 Inverse Derivative from Tabular Data
Tabular analysis
Selected values of a strictly increasing differentiable function \(f\) and its derivative \(f'\) are shown in the table. Let \(g(x) = f^{-1}(x)\). Find the value of \(g'(5)\).
| \(x\) |
\(-1\) |
\(2\) |
\(5\) |
| \(f(x)\) |
\(2\) |
\(5\) |
\(11\) |
| \(f'(x)\) |
\(4\) |
\(3\) |
\(\frac{1}{2}\) |
Final Answer \(g'(5) =\):
4 Inverse Derivative of an Algebraic Function
Analytic calculation
Let \(f(x) = 2x^3 + 3x + 1\). Given that \(f(1) = 6\), determine the exact value of \((f^{-1})'(6)\). Show all supporting work.
Final Answer \((f^{-1})'(6) =\):
Cut along line for two half-sheet copies
f⁻¹
Calculus Derivative Drill
Back • Derivatives of Inverse Functions
Rule: \((f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}\)
3 Inverse Derivative from Tabular Data
Tabular analysis
Selected values of a strictly increasing differentiable function \(f\) and its derivative \(f'\) are shown in the table. Let \(g(x) = f^{-1}(x)\). Find the value of \(g'(5)\).
| \(x\) |
\(-1\) |
\(2\) |
\(5\) |
| \(f(x)\) |
\(2\) |
\(5\) |
\(11\) |
| \(f'(x)\) |
\(4\) |
\(3\) |
\(\frac{1}{2}\) |
Final Answer \(g'(5) =\):
4 Inverse Derivative of an Algebraic Function
Analytic calculation
Let \(f(x) = 2x^3 + 3x + 1\). Given that \(f(1) = 6\), determine the exact value of \((f^{-1})'(6)\). Show all supporting work.
Final Answer \((f^{-1})'(6) =\):
Derivative Rules Answer Key
KEY
Calculus Derivative Drill • Answer Key
Teacher Guide
Complete solutions, AP scoring breakdown, and common student pitfalls
Total 4 Pts
Format 2 Front / 2 Back
Part I: Front Side Solutions 2 Points
1 Chain Rule with \(\arctan(u)\)
1 pt
\(y = \arctan\left(2x^3 - 5x\right)\)
Step 1: Identify Rule & Inner Function
Formula: \(\frac{d}{dx}[\arctan(u)] = \frac{u'}{1 + u^2}\)
Inner: \(u = 2x^3 - 5x \implies u' = 6x^2 - 5\)
Step 2: Apply Derivative Formula
\(\frac{dy}{dx} = \frac{6x^2 - 5}{1 + (2x^3 - 5x)^2}\)
Pitfall: Forgetting \(u'\) in the numerator, or confusing with \(\arcsin(u)\) by adding an incorrect radical.
Answer: \(\frac{dy}{dx} = \frac{6x^2 - 5}{1 + (2x^3 - 5x)^2}\)
2 Absolute Value Trig Derivative
1 pt
\(f(x) = \left|\cos(2x)\right|\); find \(f'(x)\) & evaluate \(f'\left(\frac{\pi}{6}\right)\)
Step 1: Apply \(\frac{d}{dx}|u| = \frac{u}{|u|} \cdot u'\)
\(u = \cos(2x) \implies u' = -2\sin(2x)\)
\(f'(x) = \frac{\cos(2x)}{|\cos(2x)|} \cdot (-2\sin(2x)) = \frac{-2\cos(2x)\sin(2x)}{|\cos(2x)|}\)
Step 2: Evaluate at \(x = \frac{\pi}{6}\)
\(2(\frac{\pi}{6}) = \frac{\pi}{3}\). Since \(\cos(\frac{\pi}{3}) = \frac{1}{2} > 0\), \(\frac{\cos(2x)}{|\cos(2x)|} = +1\).
\(f'(\frac{\pi}{6}) = (1) \cdot (-2\sin(\frac{\pi}{3})) = -2(\frac{\sqrt{3}}{2}) = -\sqrt{3}\)
Pitfall: Writing \(f'(x) = |-2\sin(2x)|\), falsely assuming absolute value derivatives must always be positive.
Answer: \(f'(x) = \frac{-2\cos(2x)\sin(2x)}{|\cos(2x)|}\); \(-\sqrt{3}\)
Part II: Back Side Solutions 2 Points
3 Inverse Derivative from Table
1 pt
\(g(x) = f^{-1}(x)\); strictly increasing \(f\); find \(g'(5)\)
Step 1: State Inverse Derivative Rule
\(g'(5) = (f^{-1})'(5) = \frac{1}{f'(f^{-1}(5))}\)
Step 2: Read Value of \(f^{-1}(5)\)
From table: \(f(2) = 5 \implies f^{-1}(5) = 2\).
Step 3: Evaluate Derivative at \(x = 2\)
From table: \(f'(2) = 3 \implies g'(5) = \frac{1}{f'(2)} = \frac{1}{3}\).
Pitfall: Evaluating at \(x = 5\) rather than \(f(x) = 5\), mistakenly computing \(\frac{1}{f'(5)} = \frac{1}{1/2} = 2\).
Answer: \(g'(5) = \frac{1}{3}\)
4 Algebraic Inverse Derivative
1 pt
\(f(x) = 2x^3 + 3x + 1\); \(f(1) = 6\); find \((f^{-1})'(6)\)
Step 1: Set Up Inverse Relationship
\(f(1) = 6 \implies f^{-1}(6) = 1\)
\((f^{-1})'(6) = \frac{1}{f'(f^{-1}(6))} = \frac{1}{f'(1)}\)
Step 2: Differentiate \(f(x)\) & Invert
\(f'(x) = 6x^2 + 3 \implies f'(1) = 6(1)^2 + 3 = 9\)
\((f^{-1})'(6) = \frac{1}{9}\)
Pitfall: Trying to invert the cubic function algebraically, or substituting \(x = 6\) into \(f'(x)\) yielding \(\frac{1}{219}\).
Answer: \((f^{-1})'(6) = \frac{1}{9}\)
Grading Rubric: 0.5 pt for correct chain/inverse rule setup; 0.5 pt for accurate algebraic evaluation.
Q1: \(\frac{6x^2-5}{1+(2x^3-5x)^2}\) Q2: \(-\sqrt{3}\) Q3: \(\frac{1}{3}\) Q4: \(\frac{1}{9}\)
