Calculus Worksheets | Differentiation Rules Worksheets - Free Printable

To solve this worksheet, we are asked to find the derivative of the inverse function, $ f^{-1}(x) $, using the formula:

$$
(f^{-1})'(x) = \frac{1}{\frac{dx}{dy}} \quad \text{where} \quad y = f^{-1}(x)
$$

This is equivalent to:

$$
(f^{-1})'(x) = \frac{1}{f'(y)} \quad \text{since} \quad \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{f'(y)}
$$

So, the general method is:
1. Find $ f'(x) $, the derivative of the original function.
2. Then, $ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} $

But since we don’t have an explicit expression for $ f^{-1}(x) $, we can only express $ (f^{-1})'(x) $ in terms of $ y $ such that $ y = f^{-1}(x) $. So:

$$
(f^{-1})'(x) = \frac{1}{f'(y)} \quad \text{where} \quad x = f(y)
$$

Let’s go through each problem step by step.

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Problem 1: $ f(x) = x^3 - x^2 - 12x $

Step 1: Compute $ f'(x) $
$$
f'(x) = 3x^2 - 2x - 12
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{3y^2 - 2y - 12}, \quad \text{where } y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{3y^2 - 2y - 12}, \quad y = f^{-1}(x)
$$

---

Problem 2: $ f(x) = x^2 - 6x + 5 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 2x - 6
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{2y - 6}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{2y - 6}, \quad y = f^{-1}(x)
$$

---

Problem 3: $ f(x) = x^5 - 15x^4 + 74x^3 - 120x^2 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 5x^4 - 60x^3 + 222x^2 - 240x
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{5y^4 - 60y^3 + 222y^2 - 240y}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{5y^4 - 60y^3 + 222y^2 - 240y}, \quad y = f^{-1}(x)
$$

---

Problem 4: $ f(x) = x^4 - 6x^3 + 9x^2 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 4x^3 - 18x^2 + 18x
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{4y^3 - 18y^2 + 18y}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{4y^3 - 18y^2 + 18y}, \quad y = f^{-1}(x)
$$

---

Problem 5: $ f(x) = -x^4 + 6x^3 $

Step 1: Compute $ f'(x) $
$$
f'(x) = -4x^3 + 18x^2
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{-4y^3 + 18y^2}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{-4y^3 + 18y^2}, \quad y = f^{-1}(x)
$$

---

Problem 6: $ f(x) = x^3 - 4x^2 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 3x^2 - 8x
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{3y^2 - 8y}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{3y^2 - 8y}, \quad y = f^{-1}(x)
$$

---

Problem 7: $ f(x) = x^4 - 4x^3 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 4x^3 - 12x^2
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{4y^3 - 12y^2}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{4y^3 - 12y^2}, \quad y = f^{-1}(x)
$$

---

Problem 8: $ f(x) = -x^5 + 11x^4 - 38x^3 + 40x^2 $

Step 1: Compute $ f'(x) $
$$
f'(x) = -5x^4 + 44x^3 - 114x^2 + 80x
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{-5y^4 + 44y^3 - 114y^2 + 80y}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{-5y^4 + 44y^3 - 114y^2 + 80y}, \quad y = f^{-1}(x)
$$

---

Problem 9: $ f(x) = x^2 - 3x - 18 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 2x - 3
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{2y - 3}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{2y - 3}, \quad y = f^{-1}(x)
$$

---

Problem 10: $ f(x) = x^3 - 4x^2 - 15x + 18 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 3x^2 - 8x - 15
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{3y^2 - 8y - 15}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{3y^2 - 8y - 15}, \quad y = f^{-1}(x)
$$

---

Problem 11: $ f(x) = x^2 - 5x + 6 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 2x - 5
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{2y - 5}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{2y - 5}, \quad y = f^{-1}(x)
$$

---

Problem 12: $ f(x) = x^5 - 13x^4 + 52x^3 - 60x^2 $

Step 1: Compute $ f'(x) $
$$
f'(x) = 5x^4 - 52x^3 + 156x^2 - 120x
$$

Then,
$$
(f^{-1})'(x) = \frac{1}{f'(y)} = \frac{1}{5y^4 - 52y^3 + 156y^2 - 120y}, \quad y = f^{-1}(x)
$$

✔ Final Answer:
$$
(f^{-1})'(x) = \frac{1}{5y^4 - 52y^3 + 156y^2 - 120y}, \quad y = f^{-1}(x)
$$

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✔ Summary of All Answers:

| Problem | $ f^{-1}'(x) $ |
|--------|------------------|
| 1 | $ \frac{1}{3y^2 - 2y - 12} $ |
| 2 | $ \frac{1}{2y - 6} $ |
| 3 | $ \frac{1}{5y^4 - 60y^3 + 222y^2 - 240y} $ |
| 4 | $ \frac{1}{4y^3 - 18y^2 + 18y} $ |
| 5 | $ \frac{1}{-4y^3 + 18y^2} $ |
| 6 | $ \frac{1}{3y^2 - 8y} $ |
| 7 | $ \frac{1}{4y^3 - 12y^2} $ |
| 8 | $ \frac{1}{-5y^4 + 44y^3 - 114y^2 + 80y} $ |
| 9 | $ \frac{1}{2y - 3} $ |
| 10 | $ \frac{1}{3y^2 - 8y - 15} $ |
| 11 | $ \frac{1}{2y - 5} $ |
| 12 | $ \frac{1}{5y^4 - 52y^3 + 156y^2 - 120y} $ |

> ⚠️ Note: In all cases, $ y = f^{-1}(x) $. Since we don't have a closed-form expression for $ f^{-1}(x) $, we leave the answer in terms of $ y $. This is standard when finding derivatives of inverse functions without explicitly solving for the inverse.

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🔍 Explanation:

The key idea is based on the formula for the derivative of an inverse function:

$$
(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
$$

So you:
1. Differentiate $ f(x) $ → get $ f'(x) $
2. Plug in $ y = f^{-1}(x) $ into $ f'(x) $
3. Take reciprocal

You do not need to find $ f^{-1}(x) $ explicitly — just use $ y $ as a placeholder.

This method works as long as $ f $ is differentiable and one-to-one (so inverse exists), and $ f'(y) \ne 0 $.

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Let me know if you'd like help simplifying any specific one or evaluating at a point!