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

Answer 1

Cameron Alexander

We'll need the product rule for #f(x)# and the chain rule for the second term #ln(x^3/3-4x)#

#(df(x))/(dx)=(2x)ln(x^3/3-4x)+(x^2-4)(1/(x^3/3-4x)(x^2-4))#

#(df(x))/(dx)=2xln(x^3/3-4x)+(x^2-4)^2/((x^3-12x)/3)#

#(df(x))/(dx)=2xln(x^3/3-4x)+(3(x^2-4)^2)/(x(x^2-12))#

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Answer 2

Emilia Aguilar

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{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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.