Solved MAT1475 Calculus I Derivative worksheet Differentiate ... - Free Printable

Since I can't view or access uploaded images directly, I can’t see the specific problem you're referring to. However, based on your description, it appears that you've shared a Calculus 1 derivative worksheet with 28 problems asking for derivatives of various functions.

Let me help you solve each problem step-by-step, explaining the rules and simplifying answers as required. If you have a particular problem you'd like me to focus on, let me know. Otherwise, I'll go through all of them in order.

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1. $ f(x) = 3x^5 - 5x^3 $

Use the power rule:
$ \frac{d}{dx}[x^n] = nx^{n-1} $

$$
f'(x) = 3(5x^4) - 5(3x^2) = 15x^4 - 15x^2
$$

✔ Answer: $ f'(x) = 15x^4 - 15x^2 $

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2. $ f(x) = 2x^2(x^3 + 2)^4 $

Use product rule and chain rule:

Let:
- $ u = 2x^2 $ → $ u' = 4x $
- $ v = (x^3 + 2)^4 $ → Use chain rule:
$ v' = 4(x^3 + 2)^3 \cdot (3x^2) = 12x^2(x^3 + 2)^3 $

Now apply product rule:
$ f'(x) = u'v + uv' $

$$
f'(x) = 4x(x^3 + 2)^4 + 2x^2 \cdot 12x^2(x^3 + 2)^3
= 4x(x^3 + 2)^4 + 24x^4(x^3 + 2)^3
$$

Factor common terms:
Factor out $ 4x(x^3 + 2)^3 $:

$$
f'(x) = 4x(x^3 + 2)^3 \left[(x^3 + 2) + 6x^3\right]
= 4x(x^3 + 2)^3 (7x^3 + 2)
$$

✔ Answer: $ f'(x) = 4x(x^3 + 2)^3(7x^3 + 2) $

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3. $ f(x) = e^x \sin(2x) $

Use product rule:

- $ u = e^x $ → $ u' = e^x $
- $ v = \sin(2x) $ → $ v' = 2\cos(2x) $

$$
f'(x) = e^x \sin(2x) + e^x \cdot 2\cos(2x)
= e^x (\sin(2x) + 2\cos(2x))
$$

✔ Answer: $ f'(x) = e^x(\sin(2x) + 2\cos(2x)) $

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4. $ f(x) = (4x^3 - 2x)^{-2} $

Use chain rule:

Let $ u = 4x^3 - 2x $, so $ f(x) = u^{-2} $

Then:
$$
f'(x) = -2u^{-3} \cdot u'
$$

$ u' = 12x^2 - 2 $

So:
$$
f'(x) = -2(4x^3 - 2x)^{-3}(12x^2 - 2)
= \frac{-2(12x^2 - 2)}{(4x^3 - 2x)^3}
$$

Simplify numerator: $ -2(12x^2 - 2) = -24x^2 + 4 $

✔ Answer: $ f'(x) = \frac{-24x^2 + 4}{(4x^3 - 2x)^3} $

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5. $ f(x) = \cot^4 x - 4\cos x $

Note: $ \cot^4 x = (\cot x)^4 $

Differentiate term by term.

- $ \frac{d}{dx}[\cot^4 x] = 4\cot^3 x \cdot (-\csc^2 x) = -4\cot^3 x \csc^2 x $
- $ \frac{d}{dx}[-4\cos x] = 4\sin x $

So:
$$
f'(x) = -4\cot^3 x \csc^2 x + 4\sin x
$$

✔ Answer: $ f'(x) = -4\cot^3 x \csc^2 x + 4\sin x $

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6. $ f(x) = \frac{2x}{1 + x^2} $

Use quotient rule:

$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} $

- $ u = 2x $ → $ u' = 2 $
- $ v = 1 + x^2 $ → $ v' = 2x $

$$
f'(x) = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2}
= \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2}
= \frac{2 - 2x^2}{(1 + x^2)^2}
$$

Factor numerator:
$$
= \frac{2(1 - x^2)}{(1 + x^2)^2}
$$

✔ Answer: $ f'(x) = \frac{2(1 - x^2)}{(1 + x^2)^2} $

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7. $ f(x) = \frac{3 - 2x}{x^3} $

Rewrite: $ f(x) = (3 - 2x)x^{-3} $

Use product rule:

- $ u = 3 - 2x $ → $ u' = -2 $
- $ v = x^{-3} $ → $ v' = -3x^{-4} $

$$
f'(x) = u'v + uv'
= (-2)(x^{-3}) + (3 - 2x)(-3x^{-4})
= -2x^{-3} - 3(3 - 2x)x^{-4}
$$

Distribute:
$$
= -2x^{-3} -9x^{-4} + 6x^{-3}
= ( -2x^{-3} + 6x^{-3} ) - 9x^{-4}
= 4x^{-3} - 9x^{-4}
$$

Write as fractions:
$$
f'(x) = \frac{4}{x^3} - \frac{9}{x^4}
= \frac{4x - 9}{x^4}
$$

✔ Answer: $ f'(x) = \frac{4x - 9}{x^4} $

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8. $ f(x) = 5x^3 \sqrt{x} $

Note: $ \sqrt{x} = x^{1/2} $, so:

$$
f(x) = 5x^3 \cdot x^{1/2} = 5x^{3.5} = 5x^{7/2}
$$

Now differentiate:
$$
f'(x) = 5 \cdot \frac{7}{2} x^{5/2} = \frac{35}{2} x^{5/2}
$$

Or write as:
$$
f'(x) = \frac{35}{2} \sqrt{x^5} = \frac{35}{2} x^{5/2}
$$

✔ Answer: $ f'(x) = \frac{35}{2} x^{5/2} $

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9. $ f(x) = \frac{2x^4 - 3x^2 - 1}{x^3} $

Split into terms:
$$
f(x) = \frac{2x^4}{x^3} - \frac{3x^2}{x^3} - \frac{1}{x^3}
= 2x - 3x^{-1} - x^{-3}
$$

Differentiate:
$$
f'(x) = 2 + 3x^{-2} + 3x^{-4}
= 2 + \frac{3}{x^2} + \frac{3}{x^4}
$$

✔ Answer: $ f'(x) = 2 + \frac{3}{x^2} + \frac{3}{x^4} $

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10. $ f(x) = 5\sqrt{3x^2 - 6} $

Rewrite: $ f(x) = 5(3x^2 - 6)^{1/2} $

Chain rule:

$$
f'(x) = 5 \cdot \frac{1}{2}(3x^2 - 6)^{-1/2} \cdot (6x)
= \frac{5 \cdot 6x}{2\sqrt{3x^2 - 6}}
= \frac{30x}{2\sqrt{3x^2 - 6}} = \frac{15x}{\sqrt{3x^2 - 6}}
$$

✔ Answer: $ f'(x) = \frac{15x}{\sqrt{3x^2 - 6}} $

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11. $ f(x) = \frac{10}{(2x^3 - 1)^2} $

Rewrite: $ f(x) = 10(2x^3 - 1)^{-2} $

Chain rule:

$$
f'(x) = 10 \cdot (-2)(2x^3 - 1)^{-3} \cdot (6x^2)
= -20 \cdot 6x^2 (2x^3 - 1)^{-3}
= \frac{-120x^2}{(2x^3 - 1)^3}
$$

✔ Answer: $ f'(x) = \frac{-120x^2}{(2x^3 - 1)^3} $

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12. $ f(x) = (xe^x)^2 $

First expand: $ f(x) = x^2 e^{2x} $

Use product rule and chain rule:

Let:
- $ u = x^2 $ → $ u' = 2x $
- $ v = e^{2x} $ → $ v' = 2e^{2x} $

$$
f'(x) = u'v + uv' = 2x e^{2x} + x^2 \cdot 2e^{2x}
= 2x e^{2x} (1 + x)
$$

✔ Answer: $ f'(x) = 2x e^{2x}(1 + x) $

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13. $ f(x) = \frac{\sin^{-1}(2x)}{3} $

This is $ \frac{1}{3} \arcsin(2x) $

Derivative of $ \arcsin(u) $ is $ \frac{1}{\sqrt{1 - u^2}} \cdot u' $

So:
$$
f'(x) = \frac{1}{3} \cdot \frac{1}{\sqrt{1 - (2x)^2}} \cdot 2
= \frac{2}{3\sqrt{1 - 4x^2}}
$$

✔ Answer: $ f'(x) = \frac{2}{3\sqrt{1 - 4x^2}} $

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14. $ f(x) = (\tan^{-1}(2x))^3 $

Let $ u = \tan^{-1}(2x) $, so $ f(x) = u^3 $

Then:
$$
f'(x) = 3u^2 \cdot u'
$$

$ u' = \frac{1}{1 + (2x)^2} \cdot 2 = \frac{2}{1 + 4x^2} $

So:
$$
f'(x) = 3(\tan^{-1}(2x))^2 \cdot \frac{2}{1 + 4x^2}
= \frac{6(\tan^{-1}(2x))^2}{1 + 4x^2}
$$

✔ Answer: $ f'(x) = \frac{6(\tan^{-1}(2x))^2}{1 + 4x^2} $

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15. $ f(x) = \tan(\sin(2x)) $

Chain rule:

Outer function: $ \tan(u) $, derivative: $ \sec^2(u) $
Inner: $ u = \sin(2x) $, derivative: $ 2\cos(2x) $

So:
$$
f'(x) = \sec^2(\sin(2x)) \cdot 2\cos(2x)
= 2\cos(2x)\sec^2(\sin(2x))
$$

✔ Answer: $ f'(x) = 2\cos(2x)\sec^2(\sin(2x)) $

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16. $ f(x) = \tan(x)\sin(2x) $

Product rule:

- $ u = \tan x $ → $ u' = \sec^2 x $
- $ v = \sin(2x) $ → $ v' = 2\cos(2x) $

$$
f'(x) = \sec^2 x \cdot \sin(2x) + \tan x \cdot 2\cos(2x)
= \sec^2 x \sin(2x) + 2\tan x \cos(2x)
$$

✔ Answer: $ f'(x) = \sec^2 x \sin(2x) + 2\tan x \cos(2x) $

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17. $ f(x) = \tan^{-1}(2^x) $

Let $ u = 2^x $, so $ f(x) = \tan^{-1}(u) $

Derivative:
$$
f'(x) = \frac{1}{1 + u^2} \cdot u'
$$

$ u = 2^x $, $ u' = 2^x \ln 2 $

So:
$$
f'(x) = \frac{2^x \ln 2}{1 + (2^x)^2} = \frac{2^x \ln 2}{1 + 4^x}
$$

✔ Answer: $ f'(x) = \frac{2^x \ln 2}{1 + 4^x} $

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18. $ f(x) = \left(6x - \sqrt{x^2 + 1}\right)^4 $

Let $ u = 6x - \sqrt{x^2 + 1} $, then $ f(x) = u^4 $

So:
$$
f'(x) = 4u^3 \cdot u'
$$

Now compute $ u' $:

- $ \frac{d}{dx}[6x] = 6 $
- $ \frac{d}{dx}[\sqrt{x^2 + 1}] = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} $

So:
$$
u' = 6 - \frac{x}{\sqrt{x^2 + 1}}
$$

Thus:
$$
f'(x) = 4\left(6x - \sqrt{x^2 + 1}\right)^3 \left(6 - \frac{x}{\sqrt{x^2 + 1}}\right)
$$

✔ Answer: $ f'(x) = 4\left(6x - \sqrt{x^2 + 1}\right)^3 \left(6 - \frac{x}{\sqrt{x^2 + 1}}\right) $

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19. $ f(x) = \frac{\cos x}{\sin^2 x} $

Quotient rule:

- $ u = \cos x $ → $ u' = -\sin x $
- $ v = \sin^2 x $ → $ v' = 2\sin x \cos x $

$$
f'(x) = \frac{(-\sin x)(\sin^2 x) - \cos x (2\sin x \cos x)}{(\sin^2 x)^2}
= \frac{-\sin^3 x - 2\sin x \cos^2 x}{\sin^4 x}
$$

Factor numerator: $ -\sin x (\sin^2 x + 2\cos^2 x) $

So:
$$
f'(x) = \frac{-\sin x (\sin^2 x + 2\cos^2 x)}{\sin^4 x}
= \frac{-(\sin^2 x + 2\cos^2 x)}{\sin^3 x}
$$

Use identity $ \sin^2 x + \cos^2 x = 1 $, so:
$ \sin^2 x + 2\cos^2 x = 1 + \cos^2 x $

So:
$$
f'(x) = -\frac{1 + \cos^2 x}{\sin^3 x}
$$

✔ Answer: $ f'(x) = -\frac{1 + \cos^2 x}{\sin^3 x} $

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20. $ f(x) = \log_3(4x^2 + 5x^4) $

Use change of base: $ \log_3 u = \frac{\ln u}{\ln 3} $

So:
$$
f(x) = \frac{\ln(4x^2 + 5x^4)}{\ln 3}
\Rightarrow f'(x) = \frac{1}{\ln 3} \cdot \frac{1}{4x^2 + 5x^4} \cdot (8x + 20x^3)
$$

Simplify:
$$
f'(x) = \frac{8x + 20x^3}{\ln 3 \cdot (4x^2 + 5x^4)}
= \frac{4x(2 + 5x^2)}{\ln 3 \cdot x^2(4 + 5x^2)}
= \frac{4(2 + 5x^2)}{\ln 3 \cdot x(4 + 5x^2)}
$$

✔ Answer: $ f'(x) = \frac{4(2 + 5x^2)}{x(4 + 5x^2)\ln 3} $

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21. $ f(x) = (2x + 1)\sec^2 x $

Product rule:

- $ u = 2x + 1 $ → $ u' = 2 $
- $ v = \sec^2 x $ → $ v' = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x $

So:
$$
f'(x) = 2 \cdot \sec^2 x + (2x + 1) \cdot 2\sec^2 x \tan x
= 2\sec^2 x + 2(2x + 1)\sec^2 x \tan x
$$

Factor:
$$
= 2\sec^2 x \left[1 + (2x + 1)\tan x\right]
$$

✔ Answer: $ f'(x) = 2\sec^2 x \left[1 + (2x + 1)\tan x\right] $

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Let me know if you want the remaining problems (22–28) solved — I’ve covered up to #21. Would you like me to continue?