How do you differentiate #f(x)=x^-3ln(1/x)+4ln(x^4)#?

Question

Isaiah Allen · Accepted Answer

Because #ln(1/x)=-lnx# and #lnx^4=4*lnx# we have that #f(x)=-x^-3*lnx+16lnx# hence its derivative is #df(x)/dx=3*x^-4*lnx-x^-3*(1/x)+16*(1/x)=>
df(x)/dx=3*x^-4*lnx-x^-4+16*x^-1#

Cameron Alexander · Answer

undefined To differentiate the function $ f(x) = x^{-3} \ln(1/x) + 4 \ln(x^4) $, apply the product rule and chain rule where necessary:

1. For the first term $ x^{-3} \ln(1/x) $:
   - Apply the product rule: $ \frac{d}{dx}(u \cdot v) = u'v + uv' $.
   - Let $ u = x^{-3} $ and $ v = \ln(1/x) $.
   - Calculate the derivatives: $ u' = -3x^{-4} $ and $ v' = \frac{-1}{x(1/x)} = -1 $.
   - Apply the product rule: $ \frac{d}{dx}(u \cdot v) = -3x^{-4} \ln(1/x) + x^{-3} (-1) $.

2. For the second term $ 4 \ln(x^4) $:
   - Apply the chain rule: $ \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} $.
   - Let $ u = x^4 $.
   - Calculate the derivative: $ \frac{du}{dx} = 4x^3 $.
   - Apply the chain rule: $ \frac{d}{dx}(\ln(x^4)) = \frac{1}{x^4} \cdot 4x^3 = 4 $.

Combine the derivatives of both terms to get the overall derivative of $ f(x) $:
$$ f'(x) = -3x^{-4} \ln(1/x) - x^{-3} + 4 $$

Simplify further if needed.