Quiz worksheet from Quizizz focusing on derivatives of logarithmic functions, featuring six multiple-choice questions with mathematical expressions and answer options.

Let's solve each of these derivative problems step by step. We'll use rules of differentiation, especially the chain rule, product rule, and derivative of logarithmic functions.

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Question 1:

Given: $ y = \ln(\cos x) $, find $ y' $

We apply the chain rule:
$$
y' = \frac{d}{dx}[\ln(\cos x)] = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x
$$

✔ Answer: C) $-\tan x$

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Question 2:

Given: $ y = (\ln(5x))^3 $, find $ y' $

Use chain rule:
Let $ u = \ln(5x) $, so $ y = u^3 $. Then:
$$
y' = 3u^2 \cdot u'
$$
Now compute $ u' = \frac{d}{dx}[\ln(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x} $

So:
$$
y' = 3(\ln(5x))^2 \cdot \frac{1}{x} = \frac{3(\ln(5x))^2}{x}
$$

✔ Answer: B) $ \frac{3(\ln(5x))^2}{x} $

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Question 3:

Given: $ y = 3x^2 \ln(4x - 5) $, find $ y' $

Use product rule:
Let $ u = 3x^2 $, $ v = \ln(4x - 5) $

Then:
$$
y' = u'v + uv'
$$
- $ u' = 6x $
- $ v' = \frac{1}{4x - 5} \cdot 4 = \frac{4}{4x - 5} $

So:
$$
y' = 6x \cdot \ln(4x - 5) + 3x^2 \cdot \frac{4}{4x - 5} = 6x \ln(4x - 5) + \frac{12x^2}{4x - 5}
$$

✔ Answer: A) $ 6x \ln(4x - 5) + \frac{12x^2}{4x - 5} $

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Question 4:

Given: $ y = \ln(\ln(\sin x)) $, find $ y' $

Apply chain rule twice:

Let $ u = \ln(\sin x) $, then $ y = \ln(u) $, so:
$$
y' = \frac{1}{u} \cdot u' = \frac{1}{\ln(\sin x)} \cdot \frac{d}{dx}[\ln(\sin x)]
$$

Now compute $ \frac{d}{dx}[\ln(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x $

So:
$$
y' = \frac{\cot x}{\ln(\sin x)}
$$

✔ Answer: C) $ \frac{\cot x}{\ln(\sin x)} $

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Question 5:

Given: $ y = \sqrt{1 + \log x} $, find $ y' $

Note: $ \log x $ usually means $ \log_{10} x $, but in calculus, sometimes it’s natural log — but here we assume $ \log x = \log_{10} x $, so:
$$
\log_{10} x = \frac{\ln x}{\ln 10}
$$

So:
$$
y = (1 + \log_{10} x)^{1/2} = \left(1 + \frac{\ln x}{\ln 10}\right)^{1/2}
$$

Differentiate using chain rule:
$$
y' = \frac{1}{2}(1 + \log_{10} x)^{-1/2} \cdot \frac{d}{dx}[\log_{10} x]
$$

We know:
$$
\frac{d}{dx}[\log_{10} x] = \frac{1}{x \ln 10}
$$

So:
$$
y' = \frac{1}{2\sqrt{1 + \log x}} \cdot \frac{1}{x \ln 10} = \frac{1}{2x \ln 10 \cdot \sqrt{1 + \log x}}
$$

Now look at options:

- D) $ \frac{1}{2 \ln 10 \cdot \sqrt{1 + \log x} \cdot x} $ → same as above!

✔ Answer: D)

> Note: Option C has $ \ln 10 \sqrt{1 + \log x} $ in denominator, but not multiplied by $ x $ — incorrect. So D is correct.

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Question 6:

Given: $ y = (3x - 5)\ln(2x) $, find $ y' $

Product rule again:
Let $ u = 3x - 5 $, $ v = \ln(2x) $

Then:
- $ u' = 3 $
- $ v' = \frac{1}{2x} \cdot 2 = \frac{1}{x} $

So:
$$
y' = u'v + uv' = 3 \cdot \ln(2x) + (3x - 5) \cdot \frac{1}{x}
= 3\ln(2x) + \frac{3x - 5}{x}
$$

Simplify:
$$
= 3\ln(2x) + 3 - \frac{5}{x}
$$

But since the question didn’t give options, we can't choose, but this is the derivative.

Wait — the image cuts off, but perhaps the next line says “then $ y' = $” and gives choices.

But based on what's visible, no options are shown for Q6.

However, if we had to write the answer:
$$
y' = 3\ln(2x) + \frac{3x - 5}{x}
$$

Or simplified:
$$
y' = 3\ln(2x) + 3 - \frac{5}{x}
$$

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✔ Final Answers:

| Question | Answer |
|--------|--------|
| 1 | C $-\tan x$ |
| 2 | B $ \frac{3(\ln(5x))^2}{x} $ |
| 3 | A $ 6x \ln(4x - 5) + \frac{12x^2}{4x - 5} $ |
| 4 | C $ \frac{\cot x}{\ln(\sin x)} $ |
| 5 | D $ \frac{1}{2 \ln 10 \cdot x \cdot \sqrt{1 + \log x}} $ |
| 6 | $ y' = 3\ln(2x) + \frac{3x - 5}{x} $ |

Let me know if you want the explanation for Q6 with options!