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

Question

Isaiah Allen · Accepted Answer

#f'(x)= (3x^3)/(x^3-4) + ln(x^3-4)# #f(x) = xln(x^3-4)# #f'(x) = x*d/dx ln(x^3-4) + ln(x^3-4)*1# [Product Rule] #= x* 1/(x^3-4) * (3x^2) + ln(x^3-4)# [Standard differential and Chain Rule] #= (3x^3)/(x^3-4) + ln(x^3-4)#

Ezekiel Alexander · Answer

undefined 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} $.

Christopher Allers · Answer

undefined 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} $.