Basic Derivatives Practice Worksheet | PDF | Derivative | Tangent - Free Printable

Let's solve each part of the Basic Derivatives Practice Worksheet step by step. We’ll:

1. Rewrite each function using positive exponents.
2. Differentiate using basic derivative rules (power rule, etc.).
3. Simplify the derivative with no negative exponents.

Then we'll solve parts G, H, and I — finding equations of tangent lines.

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Part A: $ y = \frac{5}{2x^3} $

- Rewrite:
$$
y = \frac{5}{2} x^{-3}
$$

- Differentiate:
Use power rule:
$$
\frac{dy}{dx} = \frac{5}{2} \cdot (-3)x^{-4} = -\frac{15}{2}x^{-4}
$$

- Simplify (no negative exponents):
$$
\frac{dy}{dx} = -\frac{15}{2x^4}
$$

---

Part B: $ y = \frac{5}{(2x)^3} $

- Rewrite:
$$
y = \frac{5}{8x^3} = \frac{5}{8}x^{-3}
$$

- Differentiate:
$$
\frac{dy}{dx} = \frac{5}{8} \cdot (-3)x^{-4} = -\frac{15}{8}x^{-4}
$$

- Simplify:
$$
\frac{dy}{dx} = -\frac{15}{8x^4}
$$

---

Part C: $ y = \frac{7}{3x^{-2}} $

Wait — this is a bit tricky. Let's simplify carefully.

Note: $ x^{-2} $ in the denominator means:
$$
y = \frac{7}{3x^{-2}} = \frac{7}{3} \cdot \frac{1}{x^{-2}} = \frac{7}{3} x^{2}
$$

Because $ \frac{1}{x^{-2}} = x^2 $

- Rewrite:
$$
y = \frac{7}{3}x^2
$$

- Differentiate:
$$
\frac{dy}{dx} = \frac{7}{3} \cdot 2x = \frac{14}{3}x
$$

- Simplify:
Already simplified:
$$
\frac{dy}{dx} = \frac{14}{3}x
$$

---

Part D: $ y = \sqrt[4]{x} $

This is the fourth root of $ x $, so:
$$
y = x^{1/4}
$$

- Rewrite:
$$
y = x^{1/4}
$$

- Differentiate:
$$
\frac{dy}{dx} = \frac{1}{4}x^{-3/4}
$$

- Simplify (no negative exponents):
$$
\frac{dy}{dx} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}
$$

---

Part E: $ y = \frac{1}{x^2} + \frac{1}{x^3} $

- Rewrite:
$$
y = x^{-2} + x^{-3}
$$

- Differentiate:
$$
\frac{dy}{dx} = -2x^{-3} - 3x^{-4}
$$

- Simplify:
$$
\frac{dy}{dx} = -\frac{2}{x^3} - \frac{3}{x^4}
$$

---

Part F: $ y = \frac{2}{\sqrt[3]{x}} - \frac{3}{x^2} $

First, rewrite radicals as exponents:

- $ \sqrt[3]{x} = x^{1/3} $, so $ \frac{1}{\sqrt[3]{x}} = x^{-1/3} $
- $ \frac{3}{x^2} = 3x^{-2} $

So:
- Rewrite:
$$
y = 2x^{-1/3} - 3x^{-2}
$$

- Differentiate:
$$
\frac{dy}{dx} = 2 \cdot \left(-\frac{1}{3}\right)x^{-4/3} - 3 \cdot (-2)x^{-3} = -\frac{2}{3}x^{-4/3} + 6x^{-3}
$$

- Simplify:
$$
\frac{dy}{dx} = -\frac{2}{3x^{4/3}} + \frac{6}{x^3}
$$

Or write with radicals:
$$
= -\frac{2}{3\sqrt[3]{x^4}} + \frac{6}{x^3}
$$

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Now for Parts G, H, I: Find equations of tangent lines.

We use the point-slope form:
$$
y - y_1 = m(x - x_1)
$$
where $ m = f'(x_1) $ at the given point.

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G. Find the equation of the line tangent to $ y = x^2 - 4 $ at $ (1, -3) $

- First, verify the point:
$ y = (1)^2 - 4 = 1 - 4 = -3 $ → yes, it lies on the curve.

- Differentiate:
$$
\frac{dy}{dx} = 2x
$$

- At $ x = 1 $:
$ m = 2(1) = 2 $

- Point-slope:
$$
y - (-3) = 2(x - 1) \Rightarrow y + 3 = 2x - 2 \Rightarrow y = 2x - 5
$$

✔ Answer: $ y = 2x - 5 $

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H. Find the equation of the line tangent to $ y = \sqrt{x} $ at $ (4, 2) $

- Verify: $ \sqrt{4} = 2 $ → yes.

- Rewrite: $ y = x^{1/2} $

- Differentiate:
$$
\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
$$

- At $ x = 4 $:
$ m = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4} $

- Point-slope:
$$
y - 2 = \frac{1}{4}(x - 4)
\Rightarrow y = \frac{1}{4}x - 1 + 2 = \frac{1}{4}x + 1
$$

✔ Answer: $ y = \frac{1}{4}x + 1 $

---

I. Find the equation of the line tangent to $ y = x^{-1} $ at $ (1, 1) $

- Verify: $ y = 1^{-1} = 1 $ → yes.

- Differentiate:
$$
\frac{dy}{dx} = -1 \cdot x^{-2} = -\frac{1}{x^2}
$$

- At $ x = 1 $:
$ m = -\frac{1}{1^2} = -1 $

- Point-slope:
$$
y - 1 = -1(x - 1) \Rightarrow y = -x + 1 + 1 = -x + 2
$$

✔ Answer: $ y = -x + 2 $

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✔ Final Answers Summary:

#### Table Fill-in:

| Function | Rewrite | Differentiate | Simplify |
|--------|--------|--------------|---------|
| A. $ y = \frac{5}{2x^3} $ | $ \frac{5}{2}x^{-3} $ | $ -\frac{15}{2}x^{-4} $ | $ -\frac{15}{2x^4} $ |
| B. $ y = \frac{5}{(2x)^3} $ | $ \frac{5}{8}x^{-3} $ | $ -\frac{15}{8}x^{-4} $ | $ -\frac{15}{8x^4} $ |
| C. $ y = \frac{7}{3x^{-2}} $ | $ \frac{7}{3}x^2 $ | $ \frac{14}{3}x $ | $ \frac{14}{3}x $ |
| D. $ y = \sqrt[4]{x} $ | $ x^{1/4} $ | $ \frac{1}{4}x^{-3/4} $ | $ \frac{1}{4x^{3/4}} $ or $ \frac{1}{4\sqrt[4]{x^3}} $ |
| E. $ y = \frac{1}{x^2} + \frac{1}{x^3} $ | $ x^{-2} + x^{-3} $ | $ -2x^{-3} - 3x^{-4} $ | $ -\frac{2}{x^3} - \frac{3}{x^4} $ |
| F. $ y = \frac{2}{\sqrt[3]{x}} - \frac{3}{x^2} $ | $ 2x^{-1/3} - 3x^{-2} $ | $ -\frac{2}{3}x^{-4/3} + 6x^{-3} $ | $ -\frac{2}{3x^{4/3}} + \frac{6}{x^3} $ |

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#### Tangent Lines:

- G. $ y = 2x - 5 $
- H. $ y = \frac{1}{4}x + 1 $
- I. $ y = -x + 2 $

Let me know if you'd like these filled into the worksheet format!