Final Answer:
This is the first derivative of the function \( f(x) = 6x^2 - 3x - 36 \).
Step-by-step explanation:
Certainly! To find the first derivative \( f'(x) \) of the function \( f(x) = 6x^2 - 3x - 36 \), we will use basic differentiation rules from calculus.
The power rule for derivatives states that if you have a function \( g(x) = ax^n \), then the derivative \( g'(x) = nax^{n-1} \). We can apply this rule to each term of \( f(x) \) separately.
So, let's differentiate \( f(x) \) term by term:
1. The first term is \( 6x^2 \). Applying the power rule, we multiply the exponent, 2, by the coefficient, 6, to get \( 12x \), and then we subtract 1 from the exponent, yielding our new term \( 12x^{(2-1)} = 12x \).
2. The second term is \( -3x \). This is also a power function with \( n = 1 \). Applying the power rule, we get \( -3 \cdot 1x^{(1-1)} = -3x^0 = -3 \). Since \( x^0 = 1 \), the term simplifies to just \( -3 \).
3. The third term is a constant, \( -36 \). The derivative of a constant is zero because a constant does not change as \( x \) changes.
Adding the derivatives of all the terms together, we get the first derivative of \( f(x) \):
\[ f'(x) = 12x - 3 \]
This is the first derivative of the function \( f(x) = 6x^2 - 3x - 36 \).