Differentiate f(x) = -4 + (2x-15)lm(3(sqrtx)-x+5)

To differentiate the given function \( f(x) = -4 + (2x - 15) \ln \left( 3\sqrt{x} - x + 5 \right) \), we will apply both the product rule and the chain rule.

### Step-by-step solution:

#### Step 1: Differentiate the constant term
The derivative of \(-4\) is simply zero:
\[
\frac{d}{dx} (-4) = 0
\]

#### Step 2: Apply the product rule to the second term
We have a product of two functions: \((2x - 15)\) and \( \ln \left( 3\sqrt{x} - x + 5 \right) \). Let:
\[
u(x) = 2x - 15
\]
\[
v(x) = \ln \left( 3\sqrt{x} - x + 5 \right)
\]
According to the product rule, \( \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).

#### Step 3: Differentiate \(u(x)\)
The derivative of \(u(x) = 2x - 15\) is:
\[
u'(x) = 2
\]

#### Step 4: Differentiate \(v(x)\)
To differentiate \( v(x) = \ln \left( 3\sqrt{x} - x + 5 \right) \), apply the chain rule. Let:
\[
g(x) = 3\sqrt{x} - x + 5
\]
Then:
\[
v'(x) = \frac{1}{g(x)} \cdot g'(x)
\]
Now, differentiate \( g(x) = 3\sqrt{x} - x + 5 \):
\[
g(x) = 3x^{1/2} - x + 5
\]
The derivative of \(g(x)\) is:
\[
g'(x) = \frac{3}{2}x^{-1/2} - 1
\]

So, we have:
\[
v'(x) = \frac{\frac{3}{2}x^{-1/2} - 1}{3\sqrt{x} - x + 5}
\]

#### Step 5: Apply the product rule
Now apply the product rule:
\[
f'(x) = u'(x) v(x) + u(x) v'(x)
\]
Substitute the values:
\[
f'(x) = 2 \ln \left( 3\sqrt{x} - x + 5 \right) + (2x - 15) \cdot \frac{\frac{3}{2}x^{-1/2} - 1}{3\sqrt{x} - x + 5}
\]

This is the derivative of the given function.

### Final Result:
\[
f'(x) = 2 \ln \left( 3\sqrt{x} - x + 5 \right) + \frac{(2x - 15) \left( \frac{3}{2}x^{-1/2} - 1 \right)}{3\sqrt{x} - x + 5}
\]

Let me know if you want any further details or clarifications!

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Here are five related questions:
1. How would you differentiate \( f(x) = x \ln(x^2 + 1) \)?
2. Can you simplify the expression \( \frac{3}{2}x^{-1/2} - 1 \) further for specific values of \(x\)?
3. What is the chain rule and when is it used in differentiation?
4. How does the product rule help when differentiating products of functions?
5. How would the derivative change if the function was \( \ln(4x - 7) \) instead?

**Tip:** When differentiating products or compositions of functions, always consider the product rule and chain rule as your primary tools.

Differentiate f(x) = -4 + (2x-15)ln(3(sqrtx)-x+5)

Step-by-step guide on differentiating the function f(x) = -4 + (2x-15)ln(3√x - x + 5) using the product rule and chain rule. The solution covers logarithmic differentiation and applying key calculus principles, suitable for advanced high school or early college students.

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