# What is the derivative of #f(x)=(x^2-4)ln(x^3/3-4x)#?

We'll need the product rule for

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To find the derivative of ( f(x) = (x^2 - 4) \ln\left(\frac{x^3}{3} - 4x\right) ), you can use the product rule and the chain rule. The derivative is:

[ f'(x) = (x^2 - 4) \frac{d}{dx} \ln\left(\frac{x^3}{3} - 4x\right) + \ln\left(\frac{x^3}{3} - 4x\right) \frac{d}{dx} (x^2 - 4) ]

[ = (x^2 - 4) \frac{1}{\frac{x^3}{3} - 4x} \cdot \left(3x^2 - 12\right) + \ln\left(\frac{x^3}{3} - 4x\right) \cdot 2x ]

[ = \frac{(x^2 - 4)(3x^2 - 12)}{\frac{x^3}{3} - 4x} + 2x \ln\left(\frac{x^3}{3} - 4x\right) ]

[ = \frac{(x^2 - 4)(3x^2 - 12)}{\frac{x^3}{3} - 4x} + 2x \ln\left(\frac{x^3}{3} - 4x\right) ]

[ = \frac{3x^4 - 12x^2 - 4x^2 + 16}{x - 4} + 2x \ln\left(\frac{x^3}{3} - 4x\right) ]

[ = \frac{3x^4 - 16x^2 + 16}{x - 4} + 2x \ln\left(\frac{x^3}{3} - 4x\right) ]

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