Alright, let's solve the given problems step-by-step.
(a) First, we need to find the first derivative of the function [tex]\( f(x) = x^3 + 3x^2 - 45x + 2 \)[/tex].
To find the first derivative, we use the power rule. The power rule states that if [tex]\( f(x) = x^n \)[/tex], then [tex]\( f'(x) = nx^{n-1} \)[/tex]. We'll apply this rule to each term in the function [tex]\( f(x) \)[/tex].
1. For the term [tex]\( x^3 \)[/tex]:
[tex]\[
\frac{d}{dx}(x^3) = 3x^2
\][/tex]
2. For the term [tex]\( 3x^2 \)[/tex]:
[tex]\[
\frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x
\][/tex]
3. For the term [tex]\( -45x \)[/tex]:
[tex]\[
\frac{d}{dx}(-45x) = -45
\][/tex]
4. For the constant term [tex]\( 2 \)[/tex]:
[tex]\[
\frac{d}{dx}(2) = 0
\][/tex]
Putting it all together, the first derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[
f'(x) = 3x^2 + 6x - 45
\][/tex]
(b) Now, we need to find the second derivative of the function [tex]\( f(x) \)[/tex]. To do this, we differentiate [tex]\( f'(x) \)[/tex]:
[tex]\[
f'(x) = 3x^2 + 6x - 45
\][/tex]
Let's apply the power rule again to each term in the first derivative [tex]\( f'(x) \)[/tex]:
1. For the term [tex]\( 3x^2 \)[/tex]:
[tex]\[
\frac{d}{dx}(3x^2) = 3 \cdot 2x = 6x
\][/tex]
2. For the term [tex]\( 6x \)[/tex]:
[tex]\[
\frac{d}{dx}(6x) = 6
\][/tex]
3. For the constant term [tex]\( -45 \)[/tex]:
[tex]\[
\frac{d}{dx}(-45) = 0
\][/tex]
Putting it all together, the second derivative [tex]\( f''(x) \)[/tex] is:
[tex]\[
f''(x) = 6x + 6
\][/tex]
So the final answers are:
[tex]\[
f'(x) = 3x^2 + 6x - 45
\][/tex]
[tex]\[
f''(x) = 6x + 6
\][/tex]
