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

Answer 1

Elijah Abram

#(dy)/(dx)=(-4(1-x)^3)/((1-x)^4)#

#f(x)=ln(1-x)^4#

Using the chain rule.

#(dy)/(dx)=(dy)/(du)xx(du)/(dx)#

let#" "u=(1-x)^4=>(du)/(dx)=-4(1-x)^3#

#y=lnu=>(du)/(dx)=1/u#

#(dy)/(dx)=(dy)/(du)xx(du)/(dx)#

#(dy)/(dx)=1/uxx-4(1-x)^3#

#(dy)/(dx)=(-4(1-x)^3)/((1-x)^4)#

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Answer 2

Emma Aldridge

To differentiate the function ( f(x) = \ln{(1-x)^4} ), you can use the chain rule along with the derivative of the natural logarithm function. Here's the process:

Apply the chain rule: [ \frac{d}{dx}[\ln{(u)}] = \frac{1}{u} \cdot \frac{du}{dx} ]
Identify the inner function: [ u = (1-x)^4 ]
Find the derivative of ( u ) with respect to ( x ): [ \frac{du}{dx} = 4(1-x)^3 \cdot (-1) = -4(1-x)^3 ]
Substitute ( u ) and ( \frac{du}{dx} ) into the chain rule formula: [ \frac{d}{dx}[\ln{(1-x)^4}] = \frac{1}{(1-x)^4} \cdot (-4(1-x)^3) ]
Simplify the expression: [ \frac{d}{dx}[\ln{(1-x)^4}] = \frac{-4(1-x)^3}{(1-x)^4} ]
Further simplify: [ \frac{d}{dx}[\ln{(1-x)^4}] = \frac{-4}{(1-x)} ]

So, the derivative of ( f(x) = \ln{(1-x)^4} ) is ( \frac{-4}{(1-x)} ).

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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.