Question

The value of the first derivative of \(f(x)=\left(3x^3+2x^2+x+1\right)^2\) evaluated at \(x=2\) is (in integer).

Hint:

For functions of the form \([g(x)]^2\), always use the chain rule: \[ \frac{d}{dx}[g(x)]^2=2g(x)g'(x) \] before substituting the value of \(x\).

Answer 1

Correct Answer: 3150

Step 1: See the structure.
The function is $f(x)=g(x)^2$ with $g(x)=3x^3+2x^2+x+1$.

Step 2: Apply the chain rule.
\[ f'(x)=2\,g(x)\,g'(x),\qquad g'(x)=9x^2+4x+1 \]

Step 3: Find g at 2.
\[ g(2)=3(8)+2(4)+2+1=24+8+2+1=35 \]

Step 4: Find g' at 2.
\[ g'(2)=9(4)+8+1=45 \]

Step 5: Multiply out.
\[ f'(2)=2(35)(45)=3150 \]
\[ \boxed{3150} \]