Derivatives of trigonometric functions worksheet with solutions, including problems 13 to 18.

Let's solve each of the given problems step by step. We are asked to find the derivatives of various trigonometric functions involving compositions and powers.

We will use the following derivative rules:

- Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
- Power Rule: $\frac{d}{dx}[u^n] = n u^{n-1} \cdot u'$
- Derivatives of Trig Functions:
- $\frac{d}{dx}[\tan u] = \sec^2 u \cdot u'$
- $\frac{d}{dx}[\sin u] = \cos u \cdot u'$
- $\frac{d}{dx}[\sec u] = \sec u \tan u \cdot u'$
- $\frac{d}{dx}[\csc u] = -\csc u \cot u \cdot u'$

---

Problem 13: $ y = \tan\left(\frac{1}{x^2}\right) $

Let $ u = \frac{1}{x^2} = x^{-2} $

Then $ y = \tan(u) $, so:

$$
\frac{dy}{dx} = \sec^2(u) \cdot \frac{du}{dx}
$$

Now compute $ \frac{du}{dx} $:

$$
\frac{du}{dx} = -2x^{-3} = -\frac{2}{x^3}
$$

So,

$$
\frac{dy}{dx} = \sec^2\left(\frac{1}{x^2}\right) \cdot \left(-\frac{2}{x^3}\right)
= -\frac{2}{x^3} \sec^2\left(\frac{1}{x^2}\right)
$$

✔ Answer:
$$
\boxed{-\frac{2}{x^3} \sec^2\left(\frac{1}{x^2}\right)}
$$

---

Problem 14: $ y = \sin(\sqrt{x}) $

Let $ u = \sqrt{x} = x^{1/2} $, so $ y = \sin(u) $

$$
\frac{dy}{dx} = \cos(u) \cdot \frac{du}{dx}
$$

$$
\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
$$

So,

$$
\frac{dy}{dx} = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{2\sqrt{x}}
$$

✔ Answer:
$$
\boxed{\frac{\cos(\sqrt{x})}{2\sqrt{x}}}
$$

---

Problem 15: $ y = \sec(x^3 + 5x) $

Let $ u = x^3 + 5x $, so $ y = \sec(u) $

$$
\frac{dy}{dx} = \sec(u)\tan(u) \cdot \frac{du}{dx}
$$

$$
\frac{du}{dx} = 3x^2 + 5
$$

So,

$$
\frac{dy}{dx} = \sec(x^3 + 5x)\tan(x^3 + 5x) \cdot (3x^2 + 5)
$$

✔ Answer:
$$
\boxed{(3x^2 + 5)\sec(x^3 + 5x)\tan(x^3 + 5x)}
$$

---

Problem 16: $ y = \tan^2(x^4) $

This means $ y = [\tan(x^4)]^2 $

Let $ u = \tan(x^4) $, then $ y = u^2 $

So,

$$
\frac{dy}{dx} = 2u \cdot \frac{du}{dx}
$$

Now find $ \frac{du}{dx} $. Since $ u = \tan(v) $, $ v = x^4 $

$$
\frac{du}{dx} = \sec^2(v) \cdot \frac{dv}{dx} = \sec^2(x^4) \cdot (4x^3)
$$

So,

$$
\frac{dy}{dx} = 2\tan(x^4) \cdot [4x^3 \sec^2(x^4)]
= 8x^3 \tan(x^4) \sec^2(x^4)
$$

✔ Answer:
$$
\boxed{8x^3 \tan(x^4) \sec^2(x^4)}
$$

---

Problem 17: $ y = \sec^3\left(\frac{7}{x}\right) $

This means $ y = \left[\sec\left(\frac{7}{x}\right)\right]^3 $

Let $ u = \sec\left(\frac{7}{x}\right) $, so $ y = u^3 $

$$
\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}
$$

Now compute $ \frac{du}{dx} $. Let $ v = \frac{7}{x} = 7x^{-1} $, so $ u = \sec(v) $

$$
\frac{du}{dx} = \sec(v)\tan(v) \cdot \frac{dv}{dx}
$$

$$
\frac{dv}{dx} = -\frac{7}{x^2}
$$

So,

$$
\frac{du}{dx} = \sec\left(\frac{7}{x}\right)\tan\left(\frac{7}{x}\right) \cdot \left(-\frac{7}{x^2}\right)
$$

Now plug back into $ \frac{dy}{dx} $:

$$
\frac{dy}{dx} = 3\left[\sec\left(\frac{7}{x}\right)\right]^2 \cdot \left[ -\frac{7}{x^2} \sec\left(\frac{7}{x}\right)\tan\left(\frac{7}{x}\right) \right]
$$

$$
= -\frac{21}{x^2} \sec^3\left(\frac{7}{x}\right)\tan\left(\frac{7}{x}\right)
$$

✔ Answer:
$$
\boxed{-\frac{21}{x^2} \sec^3\left(\frac{7}{x}\right)\tan\left(\frac{7}{x}\right)}
$$

---

Problem 18: $ y = \csc^2\left(\frac{x}{9}\right) $

This means $ y = \left[\csc\left(\frac{x}{9}\right)\right]^2 $

Let $ u = \csc\left(\frac{x}{9}\right) $, so $ y = u^2 $

$$
\frac{dy}{dx} = 2u \cdot \frac{du}{dx}
$$

Now $ \frac{du}{dx} = -\csc\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right) \cdot \frac{d}{dx}\left(\frac{x}{9}\right) $

$$
\frac{d}{dx}\left(\frac{x}{9}\right) = \frac{1}{9}
$$

So,

$$
\frac{du}{dx} = -\csc\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right) \cdot \frac{1}{9}
$$

Now,

$$
\frac{dy}{dx} = 2\csc\left(\frac{x}{9}\right) \cdot \left( -\frac{1}{9} \csc\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right) \right)
$$

$$
= -\frac{2}{9} \csc^2\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right)
$$

✔ Answer:
$$
\boxed{-\frac{2}{9} \csc^2\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right)}
$$

---

✔ Final Answers Summary:

| Problem | Derivative |
|--------|-----------|
| 13 | $ -\frac{2}{x^3} \sec^2\left(\frac{1}{x^2}\right) $ |
| 14 | $ \frac{\cos(\sqrt{x})}{2\sqrt{x}} $ |
| 15 | $ (3x^2 + 5)\sec(x^3 + 5x)\tan(x^3 + 5x) $ |
| 16 | $ 8x^3 \tan(x^4) \sec^2(x^4) $ |
| 17 | $ -\frac{21}{x^2} \sec^3\left(\frac{7}{x}\right)\tan\left(\frac{7}{x}\right) $ |
| 18 | $ -\frac{2}{9} \csc^2\left(\frac{x}{9}\right)\cot\left(\frac{x}{9}\right) $ |

Let me know if you'd like these explained in a different format or with graphs!