How do you differentiate #f(x)=(2x^3)(ln3x)(sinx) # using the product rule?

Question

Samantha Adkison · Accepted Answer

# f'(x) = 6x^2 \ ln3x \ sinx + 2x^2 \ sinx + 2x^3 \ ln3x \ cosx #

We have: # f(x)=(2x^3)(ln3x)(sinx) # We need to use the triple product rule, giving # f'(x) = (d/dx2x^3)(ln3x)(sinx) + (2x^3)(d/dxln3x)(sinx) + (2x^3)(ln3x)(d/dxsinx) # # " " = (6x^2)(ln3x)(sinx) + (2x^3)(1/(3x)d/dx3x)(sinx) + (2x^3)(ln3x)(cosx) # # " " = (6x^2)(ln3x)(sinx) + (2x^3)((3)/(3x))(sinx) + (2x^3)(ln3x)(cosx) # # " " = 6x^2 \ ln3x \ sinx + 2x^2 \ sinx + 2x^3 \ ln3x \ cosx #

Evelyn Agard · Answer

undefined To differentiate $ f(x) = (2x^3)(\ln(3x))(sin(x)) $ using the product rule, you would differentiate each term separately and then apply the product rule. Here's the step-by-step process:

1. Identify the terms: $ f(x) = u \cdot v \cdot w $, where $ u = 2x^3 $, $ v = \ln(3x) $, and $ w = \sin(x) $.
2. Differentiate each term:
   - $ u' = 6x^2 $ (derivative of $ 2x^3 $ with respect to $ x $).
   - $ v' = \frac{1}{3x} $ (derivative of $ \ln(3x) $ with respect to $ x $).
   - $ w' = \cos(x) $ (derivative of $ \sin(x) $ with respect to $ x $).
3. Apply the product rule: $ f'(x) = u'vw + uv'w + uvw' $.
4. Substitute the derivatives and original functions back into the formula:
   - $ f'(x) = (6x^2)(\ln(3x))(\sin(x)) + (2x^3)\left(\frac{1}{3x}\right)(\sin(x)) + (2x^3)(\ln(3x))(\cos(x)) $.

So, the derivative of $ f(x) = (2x^3)(\ln(3x))(\sin(x)) $ is $ f'(x) = (6x^2)(\ln(3x))(\sin(x)) + \frac{2}{3}(\sin(x)) + (2x^3)(\ln(3x))(\cos(x)) $.