Let `f(x)=x^(2)+2x-5` and `g(x)=5x+30`
If h(x) = 5f(x)- xg(x), then what is the derivative of h(x)?

Options

A. `-40`
B. `-20`
C. `-10`
D. 0

Answer

To find the derivative of the function \( h(x) = 5f(x) - xg(x) \), we will follow these steps:

1. **Define the functions**:
- Given \( f(x) = x^2 + 2x - 5 \)
- Given \( g(x) = 5x + 30 \)

2. **Substitute \( f(x) \) and \( g(x) \) into \( h(x) \)**:
\[ h(x) = 5f(x) - xg(x) \]
Substitute \( f(x) \) and \( g(x) \):
\[ h(x) = 5(x^2 + 2x - 5) - x(5x + 30) \]

3. **Simplify \( h(x) \)**:
- First, calculate \( 5f(x) \):
\[ 5f(x) = 5(x^2 + 2x - 5) = 5x^2 + 10x - 25 \]
- Next, calculate \( xg(x) \):
\[ xg(x) = x(5x + 30) = 5x^2 + 30x \]
- Now substitute these back into \( h(x) \):
\[ h(x) = (5x^2 + 10x - 25) - (5x^2 + 30x) \]
- Combine like terms:
\[ h(x) = 5x^2 + 10x - 25 - 5x^2 - 30x = -20x - 25 \]

4. **Differentiate \( h(x) \)**:
\[ h'(x) = \frac{d}{dx}(-20x - 25) \]
- The derivative of \(-20x\) is \(-20\).
- The derivative of \(-25\) is \(0\).
\[ h'(x) = -20 \]

Thus, the derivative of \( h(x) \) is \( -20 \).