Math practice worksheet on derivatives of logarithmic and exponential functions.

Since both `ExplainAnswer` and `TeachTopic` are false, I will skip those sections and proceed directly to solving the problems and providing the final answers.

I will solve each derivative problem one by one using standard differentiation rules (Power Rule, Product Rule, Quotient Rule, Chain Rule).

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Problem 1: $ y = 6^{2x} \sqrt{x} $

This is a product of two functions: $ u = 6^{2x} $ and $ v = \sqrt{x} = x^{1/2} $.

Use the Product Rule: $ y' = u'v + uv' $

First, find $ u' $:
$ u = 6^{2x} $
Using the chain rule for exponential functions: $ \frac{d}{dx}[a^{f(x)}] = a^{f(x)} \ln(a) \cdot f'(x) $
Here, $ a = 6 $, $ f(x) = 2x $, so $ f'(x) = 2 $
→ $ u' = 6^{2x} \ln(6) \cdot 2 = 2 \ln(6) \cdot 6^{2x} $

Next, find $ v' $:
$ v = x^{1/2} $
→ $ v' = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} $

Now apply the product rule:
$ y' = (2 \ln(6) \cdot 6^{2x}) \cdot \sqrt{x} + (6^{2x}) \cdot \left( \frac{1}{2\sqrt{x}} \right) $

Factor out $ 6^{2x} $:
$ y' = 6^{2x} \left( 2 \ln(6) \sqrt{x} + \frac{1}{2\sqrt{x}} \right) $

We can also write this as:
$ y' = 6^{2x} \left( \frac{4x \ln(6) + 1}{2\sqrt{x}} \right) $

Either form is acceptable. Let’s keep it factored nicely.

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Problem 2: $ y = (9^x - 5)^5 $

Use the Chain Rule.

Let $ u = 9^x - 5 $, then $ y = u^5 $

$ \frac{dy}{dx} = 5u^4 \cdot \frac{du}{dx} $

Now, $ \frac{du}{dx} = \frac{d}{dx}(9^x) - 0 = 9^x \ln(9) $

So,

$ y' = 5(9^x - 5)^4 \cdot 9^x \ln(9) $

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Problem 3: $ f(x) = \ln(x^2 + 4x) $

Use the Chain Rule.

Let $ u = x^2 + 4x $, then $ f(x) = \ln(u) $

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

Simplify:

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

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Problem 4: $ f(x) = 4x^3 \ln(2x + 6) $

Product Rule: $ u = 4x^3 $, $ v = \ln(2x + 6) $

$ u' = 12x^2 $

$ v' = \frac{1}{2x+6} \cdot 2 = \frac{2}{2x+6} = \frac{1}{x+3} $

Then:

$ f'(x) = u'v + uv' = 12x^2 \ln(2x+6) + 4x^3 \cdot \frac{1}{x+3} $

Simplify second term:

$ \frac{4x^3}{x+3} $

So,

$ f'(x) = 12x^2 \ln(2x+6) + \frac{4x^3}{x+3} $

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Problem 5: $ f(x) = \frac{28e^x}{5x^2} $

This is a quotient: $ \frac{u}{v} $, where $ u = 28e^x $, $ v = 5x^2 $

Quotient Rule: $ f'(x) = \frac{u'v - uv'}{v^2} $

$ u' = 28e^x $

$ v' = 10x $

So,

$ f'(x) = \frac{(28e^x)(5x^2) - (28e^x)(10x)}{(5x^2)^2} $

Factor numerator:

$ = \frac{28e^x (5x^2 - 10x)}{25x^4} $

Factor inside: $ 5x^2 - 10x = 5x(x - 2) $

So,

$ f'(x) = \frac{28e^x \cdot 5x(x - 2)}{25x^4} = \frac{140x e^x (x - 2)}{25x^4} $

Cancel one x:

$ = \frac{140 e^x (x - 2)}{25x^3} $

Simplify fraction: divide numerator and denominator by 5:

$ = \frac{28 e^x (x - 2)}{5x^3} $

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Problem 6: $ y = -(x^3 + 2x^4) \ln(2x + 1) $

Product Rule: $ u = -(x^3 + 2x^4) $, $ v = \ln(2x + 1) $

First, simplify $ u = -x^3 - 2x^4 $

$ u' = -3x^2 - 8x^3 $

$ v' = \frac{1}{2x+1} \cdot 2 = \frac{2}{2x+1} $

Then:

$ y' = u'v + uv' = (-3x^2 - 8x^3)\ln(2x+1) + (-x^3 - 2x^4) \cdot \frac{2}{2x+1} $

Write as:

$ y' = -(3x^2 + 8x^3)\ln(2x+1) - \frac{2x^3(1 + 2x)}{2x+1} $

Note: $ 1 + 2x = 2x + 1 $, so:

$ \frac{2x^3(2x+1)}{2x+1} = 2x^3 $, provided $ 2x+1 \ne 0 $

So the second term simplifies to $ -2x^3 $

Thus:

$ y' = -(3x^2 + 8x^3)\ln(2x+1) - 2x^3 $

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Problem 7: $ f(x) = (-e^{x^2 - 2x}) \ln(4x) $

Product Rule: $ u = -e^{x^2 - 2x} $, $ v = \ln(4x) $

First, $ u' $: use chain rule

Let $ w = x^2 - 2x $, so $ u = -e^w $, $ u' = -e^w \cdot w' = -e^{x^2 - 2x} \cdot (2x - 2) $

$ v = \ln(4x) = \ln(4) + \ln(x) $, so $ v' = \frac{1}{x} $

Alternatively, directly: $ \frac{d}{dx}[\ln(4x)] = \frac{1}{4x} \cdot 4 = \frac{1}{x} $

So,

$ f'(x) = u'v + uv' = [ -e^{x^2 - 2x} (2x - 2) ] \cdot \ln(4x) + [ -e^{x^2 - 2x} ] \cdot \frac{1}{x} $

Factor out $ -e^{x^2 - 2x} $:

$ f'(x) = -e^{x^2 - 2x} \left[ (2x - 2)\ln(4x) + \frac{1}{x} \right] $

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Problem 8: $ y = \frac{e^{3x} - x^2}{x} $

Quotient Rule: $ u = e^{3x} - x^2 $, $ v = x $

$ u' = 3e^{3x} - 2x $, $ v' = 1 $

$ y' = \frac{(3e^{3x} - 2x)(x) - (e^{3x} - x^2)(1)}{x^2} $

Expand numerator:

$ = \frac{3x e^{3x} - 2x^2 - e^{3x} + x^2}{x^2} $

Combine like terms:

$ = \frac{3x e^{3x} - e^{3x} - x^2}{x^2} $

Factor $ e^{3x} $:

$ = \frac{e^{3x}(3x - 1) - x^2}{x^2} $

Or split:

$ = \frac{e^{3x}(3x - 1)}{x^2} - 1 $

Either form is fine. Let's leave as single fraction:

$ y' = \frac{e^{3x}(3x - 1) - x^2}{x^2} $

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Problem 9: $ f(x) = \frac{xe^{x^2}}{\ln x} $

Quotient Rule: $ u = x e^{x^2} $, $ v = \ln x $

First, find $ u' $: product rule on $ x \cdot e^{x^2} $

Let $ a = x $, $ b = e^{x^2} $

$ a' = 1 $, $ b' = e^{x^2} \cdot 2x $

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

$ v' = \frac{1}{x} $

Now quotient rule:

$ f'(x) = \frac{u'v - uv'}{v^2} = \frac{ [e^{x^2}(1 + 2x^2)] \cdot \ln x - [x e^{x^2}] \cdot \frac{1}{x} }{ (\ln x)^2 } $

Simplify second term in numerator: $ x e^{x^2} \cdot \frac{1}{x} = e^{x^2} $

So:

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

Factor $ e^{x^2} $:

$ f'(x) = \frac{ e^{x^2} \left[ (1 + 2x^2) \ln x - 1 \right] }{ (\ln x)^2 } $

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Problem 10: $ f(x) = (e^x - 1)^2 (x^2 + 4)^3 $

Product Rule: $ u = (e^x - 1)^2 $, $ v = (x^2 + 4)^3 $

Find $ u' $: chain rule

$ u = [g(x)]^2 $, $ g(x) = e^x - 1 $, $ g'(x) = e^x $

So $ u' = 2(e^x - 1) \cdot e^x $

Find $ v' $: chain rule

$ v = [h(x)]^3 $, $ h(x) = x^2 + 4 $, $ h'(x) = 2x $

So $ v' = 3(x^2 + 4)^2 \cdot 2x = 6x(x^2 + 4)^2 $

Now product rule:

$ f'(x) = u'v + uv' = [2e^x(e^x - 1)] \cdot (x^2 + 4)^3 + (e^x - 1)^2 \cdot [6x(x^2 + 4)^2] $

Factor common terms: both have $ (e^x - 1) $ and $ (x^2 + 4)^2 $

So factor out $ 2(e^x - 1)(x^2 + 4)^2 $:

First term: $ 2e^x(e^x - 1)(x^2 + 4)^3 = 2(e^x - 1)(x^2 + 4)^2 \cdot e^x (x^2 + 4) $

Second term: $ 6x(e^x - 1)^2 (x^2 + 4)^2 = 2(e^x - 1)(x^2 + 4)^2 \cdot 3x(e^x - 1) $

So overall:

$ f'(x) = 2(e^x - 1)(x^2 + 4)^2 \left[ e^x(x^2 + 4) + 3x(e^x - 1) \right] $

You can leave it like this, or expand inside if needed, but this is simplified.

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Final Answers:

1. $ y' = 6^{2x} \left( 2 \ln(6) \sqrt{x} + \frac{1}{2\sqrt{x}} \right) $

2. $ y' = 5(9^x - 5)^4 \cdot 9^x \ln(9) $

3. $ f'(x) = \frac{2(x + 2)}{x(x + 4)} $

4. $ f'(x) = 12x^2 \ln(2x+6) + \frac{4x^3}{x+3} $

5. $ f'(x) = \frac{28 e^x (x - 2)}{5x^3} $

6. $ y' = -(3x^2 + 8x^3)\ln(2x+1) - 2x^3 $

7. $ f'(x) = -e^{x^2 - 2x} \left[ (2x - 2)\ln(4x) + \frac{1}{x} \right] $

8. $ y' = \frac{e^{3x}(3x - 1) - x^2}{x^2} $

9. $ f'(x) = \frac{ e^{x^2} \left[ (1 + 2x^2) \ln x - 1 \right] }{ (\ln x)^2 } $

10. $ f'(x) = 2(e^x - 1)(x^2 + 4)^2 \left[ e^x(x^2 + 4) + 3x(e^x - 1) \right] $