Worksheet with 20 problems on finding derivatives of trigonometric functions.

Let’s solve each problem one by one. We’ll use basic derivative rules for trig functions:

- d/dx[sin x] = cos x
- d/dx[cos x] = -sin x
- d/dx[tan x] = sec²x
- d/dx[cot x] = -csc²x
- d/dx[sec x] = sec x tan x
- d/dx[csc x] = -csc x cot x

Also, we’ll need the chain rule and product rule when needed.

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1. y = 3 sin x + 3 cos x – tan x

Derivative term by term:
- d/dx[3 sin x] = 3 cos x
- d/dx[3 cos x] = 3(-sin x) = -3 sin x
- d/dx[-tan x] = -sec²x

So:
y’ = 3 cos x – 3 sin x – sec²x

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2. y = 2 cos²x

This is 2*(cos x)^2 → use chain rule.

Let u = cos x → y = 2u² → dy/du = 4u, du/dx = -sin x

So:
dy/dx = 4u * (-sin x) = 4 cos x * (-sin x) = -4 sin x cos x

(You can also write as -2 sin(2x), but we’ll leave it as is.)

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3. y = tan²x + sec²x

First term: tan²x → let u = tan x → y = u² → dy/dx = 2u * sec²x = 2 tan x sec²x

Second term: sec²x → let u = sec x → y = u² → dy/dx = 2u * (sec x tan x) = 2 sec x * sec x tan x = 2 sec²x tan x

Wait — actually, both terms have same derivative? Let me check:

d/dx[tan²x] = 2 tan x * sec²x
d/dx[sec²x] = 2 sec x * (sec x tan x) = 2 sec²x tan x

Yes! So total:
y’ = 2 tan x sec²x + 2 sec²x tan x = 4 tan x sec²x

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4. y = 1 – sin²x

Derivative of constant 1 is 0.
Derivative of -sin²x: let u = sin x → y = -u² → dy/dx = -2u * cos x = -2 sin x cos x

So:
y’ = -2 sin x cos x

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5. y = csc x

Direct formula:
y’ = -csc x cot x

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6. y = -2 cot x + 3 cot²x

First term: d/dx[-2 cot x] = -2*(-csc²x) = 2 csc²x

Second term: 3 cot²x → let u = cot x → y = 3u² → dy/dx = 6u * (-csc²x) = 6 cot x * (-csc²x) = -6 cot x csc²x

So:
y’ = 2 csc²x – 6 cot x csc²x

Factor if you want: 2 csc²x (1 – 3 cot x)

But we’ll leave expanded unless asked.

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7. y = x³ tan x

Product rule: (uv)’ = u’v + uv’

u = x³ → u’ = 3x²
v = tan x → v’ = sec²x

So:
y’ = 3x² tan x + x³ sec²x

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8. Find equation of tangent line to y = cos x at x = π/2

Step 1: Find point on curve.

At x = π/2, y = cos(π/2) = 0 → point is (π/2, 0)

Step 2: Find slope = derivative at that point.

y’ = -sin x → at x = π/2, y’ = -sin(π/2) = -1

Step 3: Use point-slope form: y – y₁ = m(x – x₁)

y – 0 = -1(x – π/2)

So:
y = -x + π/2

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9. f(x) = tan x sec²x

Product rule: u = tan x, v = sec²x

u’ = sec²x
v’ = 2 sec x * (sec x tan x) = 2 sec²x tan x [from earlier]

So:
f’(x) = u’v + uv’ = sec²x * sec²x + tan x * 2 sec²x tan x
= sec⁴x + 2 tan²x sec²x

We can factor sec²x:
f’(x) = sec²x (sec²x + 2 tan²x)

Or leave as is.

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10. f(x) = tan³x cos²(4x)

This is product of two functions: u = tan³x, v = cos²(4x)

Use product rule: f’ = u’v + uv’

First, u = tan³x → u’ = 3 tan²x * sec²x [chain rule]

v = cos²(4x) → let w = cos(4x), so v = w² → dv/dw = 2w, dw/dx = -sin(4x)*4

So dv/dx = 2 cos(4x) * (-4 sin(4x)) = -8 cos(4x) sin(4x)

Now plug in:

f’(x) = [3 tan²x sec²x] * [cos²(4x)] + [tan³x] * [-8 cos(4x) sin(4x)]

So:
f’(x) = 3 tan²x sec²x cos²(4x) – 8 tan³x cos(4x) sin(4x)

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11. f(x) = sec³x sin²x

Product rule: u = sec³x, v = sin²x

u’ = 3 sec²x * (sec x tan x) = 3 sec³x tan x
v’ = 2 sin x cos x

So:
f’(x) = u’v + uv’ = [3 sec³x tan x][sin²x] + [sec³x][2 sin x cos x]

Factor out sec³x sin x:

= sec³x sin x [3 tan x sin x + 2 cos x]

But maybe better to leave as:

f’(x) = 3 sec³x tan x sin²x + 2 sec³x sin x cos x

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12. f(x) = sin⁴√x

Write as [sin(x^{1/2})]^4

Chain rule multiple times.

Let u = x^{1/2} → du/dx = (1/2)x^{-1/2}

Let v = sin u → dv/du = cos u

Let w = v^4 → dw/dv = 4v³

So df/dx = dw/dv * dv/du * du/dx = 4v³ * cos u * (1/(2√x))

Substitute back:

= 4 [sin(√x)]³ * cos(√x) * (1/(2√x))
= (2 / √x) sin³(√x) cos(√x)

So:
f’(x) = \frac{2}{\sqrt{x}} \sin^3(\sqrt{x}) \cos(\sqrt{x})

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13. f(x) = tan⁵(x³ + x)

Outer function: u^5, inner: u = tan(v), v = x³ + x

df/dx = 5 tan⁴(v) * sec²(v) * (3x² + 1)

Where v = x³ + x

So:
f’(x) = 5 tan⁴(x³ + x) sec²(x³ + x) (3x² + 1)

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14. f(x) = sin⁻¹(√x)

Derivative of arcsin(u) is 1/√(1–u²) * u’

Here u = √x = x^{1/2}, u’ = 1/(2√x)

So:
f’(x) = 1 / √(1 – (√x)²) * (1/(2√x)) = 1 / √(1 – x) * 1/(2√x)

So:
f’(x) = \frac{1}{2\sqrt{x}\sqrt{1 - x}}

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15. y = tan⁻¹(3x – 5)

Derivative of arctan(u) is 1/(1+u²) * u’

u = 3x – 5 → u’ = 3

So:
y’ = \frac{3}{1 + (3x - 5)^2}

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16. f(x) = sin⁻¹((1/3)x)

u = (1/3)x → u’ = 1/3

f’(x) = 1 / √(1 – u²) * u’ = 1 / √(1 – (x/3)²) * (1/3)

Simplify denominator: √(1 – x²/9) = √((9 – x²)/9) = √(9 – x²)/3

So:
f’(x) = [1 / (√(9 – x²)/3)] * (1/3) = [3 / √(9 – x²)] * (1/3) = 1 / √(9 – x²)

Wait — let's do step by step:

f’(x) = \frac{1}{\sqrt{1 - (\frac{x}{3})^2}} \cdot \frac{1}{3} = \frac{1}{3 \sqrt{1 - \frac{x^2}{9}}} = \frac{1}{3 \cdot \frac{\sqrt{9 - x^2}}{3}} = \frac{1}{\sqrt{9 - x^2}}

Yes!

So:
f’(x) = \frac{1}{\sqrt{9 - x^2}}

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17. g(x) = tan⁻¹(x²)

u = x² → u’ = 2x

g’(x) = 1/(1 + u²) * u’ = 1/(1 + x⁴) * 2x

So:
g’(x) = \frac{2x}{1 + x^4}

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18. y = x² arctan(x²)

Product rule: u = x², v = arctan(x²)

u’ = 2x
v’ = 1/(1 + (x²)²) * 2x = 2x / (1 + x⁴)

So:
y’ = u’v + uv’ = 2x * arctan(x²) + x² * [2x / (1 + x⁴)]

= 2x arctan(x²) + \frac{2x^3}{1 + x^4}

So:
y’ = 2x \arctan(x^2) + \frac{2x^3}{1 + x^4}

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19. f(x) = sec⁻¹(√(x² – 1))

Derivative of arcsec(u) is 1/(|u|√(u² – 1)) * u’

Here u = √(x² – 1)

First, note domain: x² – 1 > 0 → |x| > 1

u = (x² – 1)^{1/2} → u’ = (1/2)(x² – 1)^{-1/2} * 2x = x / √(x² – 1)

Now, derivative:

f’(x) = 1 / [ |u| √(u² – 1) ] * u’

But u = √(x² – 1), so |u| = √(x² – 1) since square root is non-negative.

u² = x² – 1 → u² – 1 = x² – 2

Wait — that might be messy. Let me double-check.

Actually, standard formula: d/dx[arcsec(u)] = 1/(|u|√(u² – 1)) * du/dx

Here u = √(x² – 1)

So u² = x² – 1 → u² – 1 = x² – 2

But only if x² – 2 ≥ 0? That would restrict domain further. Maybe I made a mistake.

Wait — perhaps simplify first.

Note: sec⁻¹(√(x² – 1)) — let θ = sec⁻¹(√(x² – 1)), then sec θ = √(x² – 1)

Then tan θ = √(sec²θ – 1) = √(x² – 1 – 1) = √(x² – 2)

But this seems complicated.

Alternatively, recognize that sec⁻¹(√(x² – 1)) might relate to other inverse trig functions.

Actually, let’s compute directly.

f’(x) = 1 / [ |√(x² – 1)| * √( (√(x² – 1))² – 1 ) ] * d/dx[√(x² – 1)]

= 1 / [ √(x² – 1) * √(x² – 1 – 1) ] * [x / √(x² – 1)] (since √(x² – 1) > 0 for |x|>1)

= 1 / [ √(x² – 1) * √(x² – 2) ] * [x / √(x² – 1)]

= x / [ (x² – 1) √(x² – 2) ]

But this requires x² > 2, which may not be intended.

Wait — perhaps there’s an identity.

Recall: sec⁻¹(z) = cos⁻¹(1/z)

So f(x) = cos⁻¹(1 / √(x² – 1))

Then derivative of cos⁻¹(u) is -1/√(1 – u²) * u’

u = (x² – 1)^{-1/2}

u’ = (-1/2)(x² – 1)^{-3/2} * 2x = -x (x² – 1)^{-3/2}

Now, 1 – u² = 1 – 1/(x² – 1) = (x² – 1 – 1)/(x² – 1) = (x² – 2)/(x² – 1)

So √(1 – u²) = √(x² – 2)/√(x² – 1)

Thus:

f’(x) = -1 / [ √(x² – 2)/√(x² – 1) ] * [ -x (x² – 1)^{-3/2} ]

= [ √(x² – 1) / √(x² – 2) ] * [ x / (x² – 1)^{3/2} ]

= x / [ √(x² – 2) * (x² – 1) ]

Same as before.

So:
f’(x) = \frac{x}{(x^2 - 1)\sqrt{x^2 - 2}}

But this is valid only for |x| > √2. Perhaps the problem assumes that.

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20. y = x² sec⁻¹(5x)

Product rule: u = x², v = sec⁻¹(5x)

u’ = 2x
v’ = 1/(|5x| √((5x)² – 1)) * 5 = 5 / (|5x| √(25x² – 1)) = 1 / (|x| √(25x² – 1)) [since 5/|5x| = 1/|x|]

Assuming x > 0 for simplicity (as often done unless specified), |x| = x, so v’ = 1/(x √(25x² – 1))

Then y’ = u’v + uv’ = 2x sec⁻¹(5x) + x² * [1/(x √(25x² – 1))] = 2x sec⁻¹(5x) + x / √(25x² – 1)

So:
y’ = 2x \sec^{-1}(5x) + \frac{x}{\sqrt{25x^2 - 1}}

If x < 0, we’d have absolute value, but typically in such problems, we assume domain where expression is defined and positive.

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Now, compiling all final answers clearly:

Final Answer:
1. \( y' = 3\cos x - 3\sin x - \sec^2 x \)
2. \( y' = -4\sin x \cos x \)
3. \( y' = 4\tan x \sec^2 x \)
4. \( y' = -2\sin x \cos x \)
5. \( y' = -\csc x \cot x \)
6. \( y' = 2\csc^2 x - 6\cot x \csc^2 x \)
7. \( y' = 3x^2 \tan x + x^3 \sec^2 x \)
8. \( y = -x + \frac{\pi}{2} \)
9. \( f'(x) = \sec^4 x + 2\tan^2 x \sec^2 x \) or \( \sec^2 x (\sec^2 x + 2\tan^2 x) \)
10. \( f'(x) = 3\tan^2 x \sec^2 x \cos^2(4x) - 8\tan^3 x \cos(4x) \sin(4x) \)
11. \( f'(x) = 3\sec^3 x \tan x \sin^2 x + 2\sec^3 x \sin x \cos x \)
12. \( f'(x) = \frac{2}{\sqrt{x}} \sin^3(\sqrt{x}) \cos(\sqrt{x}) \)
13. \( f'(x) = 5\tan^4(x^3 + x) \sec^2(x^3 + x) (3x^2 + 1) \)
14. \( f'(x) = \frac{1}{2\sqrt{x}\sqrt{1 - x}} \)
15. \( y' = \frac{3}{1 + (3x - 5)^2} \)
16. \( f'(x) = \frac{1}{\sqrt{9 - x^2}} \)
17. \( g'(x) = \frac{2x}{1 + x^4} \)
18. \( y' = 2x \arctan(x^2) + \frac{2x^3}{1 + x^4} \)
19. \( f'(x) = \frac{x}{(x^2 - 1)\sqrt{x^2 - 2}} \) (for |x| > √2)
20. \( y' = 2x \sec^{-1}(5x) + \frac{x}{\sqrt{25x^2 - 1}} \) (assuming x > 0)