Question

let f(x) = 3x⁴ ln x. f(x) =

Explanation:

Step1: Find the first derivative \( f'(x) \)

We use the product rule: if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Let \( u(x) = 3x^4 \) and \( v(x) = \ln x \).

• \( u'(x) = \frac{d}{dx}(3x^4) = 12x^3 \) (using the power rule \( \frac{d}{dx}(x^n) = nx^{n - 1} \))
• \( v'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \)

So, \( f'(x) = 12x^3 \cdot \ln x + 3x^4 \cdot \frac{1}{x} = 12x^3\ln x + 3x^3 \)

Step2: Find the second derivative \( f''(x) \)

We differentiate \( f'(x) = 12x^3\ln x + 3x^3 \) term - by - term.

For the first term \( 12x^3\ln x \):

Use the product rule again. Let \( u(x)=12x^3 \) and \( v(x)=\ln x \)

• \( u'(x)=\frac{d}{dx}(12x^3)=36x^2 \)
• \( v'(x)=\frac{1}{x} \)

So, the derivative of \( 12x^3\ln x \) is \( 36x^2\ln x+12x^3\cdot\frac{1}{x}=36x^2\ln x + 12x^2 \)

For the second term \( 3x^3 \):

Using the power rule, \( \frac{d}{dx}(3x^3)=9x^2 \)

Now, add the derivatives of the two terms:
\( f''(x)=(36x^2\ln x + 12x^2)+9x^2=36x^2\ln x+21x^2 \)

Answer:

\( f''(x)=36x^{2}\ln x + 21x^{2} \)