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

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To find the derivative of ( f(x) = x \ln(x^3 - 4) ), you can use the product rule of differentiation.

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

Now, ( \frac{d}{dx}(x) = 1 ) and ( \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{d}{dx}(u) ), where ( u = x^3 - 4 ).

So,

[ f'(x) = 1 \cdot \ln(x^3 - 4) + x \cdot \frac{1}{x^3 - 4} \cdot \frac{d}{dx}(x^3 - 4) ]

Using the chain rule, ( \frac{d}{dx}(x^3 - 4) = 3x^2 ).

Substituting back,

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

Therefore, the derivative of ( f(x) = x \ln(x^3 - 4) ) with respect to ( x ) is ( f'(x) = \ln(x^3 - 4) + \frac{3x^2}{x^3 - 4} ).

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To find the derivative of ( f(x) = x \ln(x^3 - 4) ), we'll use the product rule.

Let ( u = x ) and ( v = \ln(x^3 - 4) ).

Then, ( u' = 1 ) and ( v' = \frac{1}{x^3 - 4} \cdot (3x^2) ).

Applying the product rule, we have:

[ f'(x) = u'v + uv' ]

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

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

So, the derivative of ( f(x) = x \ln(x^3 - 4) ) is ( f'(x) = \ln(x^3 - 4) + \frac{3x^3}{x^3 - 4} ).

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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.

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.

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.

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