Worksheet on finding derivatives of logarithmic functions with solutions.

Let’s solve each problem one by one. We’ll use the rules for derivatives of logarithmic functions.

Remember:
- The derivative of ln(u) is (1/u) * u’
- The derivative of log_b(u) is (1/(u * ln(b))) * u’
- Use chain rule when needed: if y = f(g(x)), then y’ = f’(g(x)) * g’(x)
- For products, use product rule: (uv)’ = u’v + uv’

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Problem 9: y = log₃(2x - 3)

Step 1: Recall derivative formula for log base b:
d/dx [log_b(u)] = (1 / (u * ln(b))) * u’

Here, u = 2x - 3 → u’ = 2
b = 3

So,
y’ = (1 / ((2x - 3) * ln(3))) * 2
→ y’ = 2 / [(2x - 3) ln(3)]

✔ Check: Chain rule applied correctly? Yes. Derivative of inside function is 2. Base 3 handled with ln(3). Good.

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Problem 10: y = log₁₀(x / (x + 1))

We can simplify first using log property:
log(a/b) = log(a) - log(b)

So, y = log₁₀(x) - log₁₀(x + 1)

Now take derivative term by term:

Derivative of log₁₀(x) = 1/(x * ln(10))
Derivative of log₁₀(x+1) = 1/((x+1) * ln(10))

So,
y’ = [1/(x ln(10))] - [1/((x+1) ln(10))]
= [1/ln(10)] * [1/x - 1/(x+1)]

Combine fractions:
1/x - 1/(x+1) = [(x+1) - x] / [x(x+1)] = 1 / [x(x+1)]

So,
y’ = 1 / [ln(10) * x(x+1)]

✔ Check: Used log properties to simplify — makes differentiation easier. Algebra correct? Yes.

Alternative method (without simplifying):
Use chain rule on log₁₀(u), where u = x/(x+1)

u’ = [(1)(x+1) - x(1)] / (x+1)² = (x+1 - x)/(x+1)² = 1/(x+1)²

Then y’ = [1/(u * ln(10))] * u’ = [1 / ( (x/(x+1)) * ln(10) ) ] * [1/(x+1)²]

= [ (x+1)/x * 1/ln(10) ] * [1/(x+1)²] = 1/[x(x+1) ln(10)] → same answer!

Good.

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Problem 11: y = ln(√x)

First, rewrite √x as x^(1/2)

So, y = ln(x^{1/2}) = (1/2) ln(x) ← using log power rule

Now derivative:
y’ = (1/2) * (1/x) = 1/(2x)

✔ Check: Log property used correctly? Yes. Derivative of ln(x) is 1/x. Simple and clean.

Alternative: Chain rule directly
y = ln(u), u = x^{1/2}, u’ = (1/2)x^{-1/2}

y’ = (1/u) * u’ = (1/√x) * (1/(2√x)) = 1/(2x) → same.

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Problem 12: y = ln(√(x² - 3))

Rewrite: √(x² - 3) = (x² - 3)^{1/2}

So, y = ln( (x² - 3)^{1/2} ) = (1/2) ln(x² - 3)

Now derivative:
y’ = (1/2) * [1/(x² - 3)] * (derivative of inside)

Inside: x² - 3 → derivative = 2x

So,
y’ = (1/2) * [1/(x² - 3)] * 2x = x / (x² - 3)

✔ Check: Power rule on log, chain rule applied. 2 and 1/2 cancel. Correct.

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Problem 13: y = √x * ln(x)

This is a product: u = √x = x^{1/2}, v = ln(x)

Product rule: y’ = u’v + uv’

u’ = (1/2)x^{-1/2} = 1/(2√x)
v’ = 1/x

So,
y’ = [1/(2√x)] * ln(x) + √x * [1/x]

Simplify second term: √x / x = 1/√x

So,
y’ = [ln(x) / (2√x)] + [1 / √x]

Get common denominator (2√x):

= [ln(x) + 2] / (2√x)

✔ Check: Product rule applied. Terms simplified correctly. Final form neat.

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Problem 14: y = x² * log₇(x² + 1)

Product rule again: u = x², v = log₇(x² + 1)

u’ = 2x

v’ = derivative of log₇(u) = [1 / (u * ln(7))] * u’
Here u = x² + 1 → u’ = 2x
So v’ = [1 / ((x² + 1) ln(7))] * 2x = 2x / [(x² + 1) ln(7)]

Now apply product rule:

y’ = u’v + uv’
= 2x * log₇(x² + 1) + x² * [2x / ((x² + 1) ln(7))]

Simplify second term:
x² * 2x = 2x³
So second term: 2x³ / [(x² + 1) ln(7)]

Final answer:
y’ = 2x log₇(x² + 1) + \frac{2x^3}{(x^2 + 1)\ln(7)}

✔ Check: Product rule done. Derivative of log base 7 handled with ln(7). Algebra correct.

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Final Answer:

9) $\frac{2}{(2x - 3)\ln 3}$

10) $\frac{1}{\ln 10 \cdot x(x + 1)}$

11) $\frac{1}{2x}$

12) $\frac{x}{x^2 - 3}$

13) $\frac{\ln x + 2}{2\sqrt{x}}$

14) $2x \log_7(x^2 + 1) + \frac{2x^3}{(x^2 + 1)\ln 7}$