To find the derivative of the function:
f(x)=(ln(x−1))2ln(x−1)−1
we will use the quotient rule. The quotient rule states that for a function of the form f(x)=v(x)u(x), the derivative f′(x) is given by:
f′(x)=(v(x))2u′(x)v(x)−u(x)v′(x).
Step 1: Define u(x) and v(x)
- u(x)=ln(x−1)−1
- v(x)=(ln(x−1))2
Step 2: Compute u′(x) and v′(x)
Derivative of u(x)
u(x)=ln(x−1)−1
Using the chain rule:
u′(x)=x−11.
Derivative of v(x)
v(x)=(ln(x−1))2
Using the chain rule and the power rule:
v′(x)=2ln(x−1)⋅x−11=x−12ln(x−1).
Step 3: Apply the Quotient Rule
Substituting u(x), v(x), u′(x), and v′(x) into the quotient rule:
f′(x)=(v(x))2u′(x)v(x)−u(x)v′(x).
Plugging in the expressions:
f′(x)=(ln(x−1)2)2(x−11)(ln(x−1))2−(ln(x−1)−1)(x−12ln(x−1)).
Step 4: Simplify the Numerator
Factor out x−11 from both terms in the numerator:
f′(x)=(ln(x−1))4x−11[(ln(x−1))2−2ln(x−1)(ln(x−1)−1)].
Simplify inside the brackets:
(ln(x−1))2−2ln(x−1)(ln(x−1)−1)=(ln(x−1))2−2(ln(x−1))2+2ln(x−1).
Combine like terms:
=−(ln(x−1))2+2ln(x−1).
Thus, the expression for f′(x) becomes:
f′(x)=(ln(x−1))4x−11[−(ln(x−1))2+2ln(x−1)].
Step 5: Final Expression
Combining the terms:
f′(x)=(x−1)(ln(x−1))4−(ln(x−1))2+2ln(x−1).
This is the derivative of the function f(x).
Do you need a more detailed explanation or simplification?
Related Questions:
- What is the derivative of g(x)=(ln(x))2ln(x)?
- How do you apply the quotient rule to f(x)=x2ex?
- What is the second derivative of f(x)=(ln(x−1))2ln(x−1)−1?
- How do you simplify expressions involving logarithmic derivatives?
- What are the critical points of f(x)=(ln(x−1))2ln(x−1)−1?
Tip:
When dealing with derivatives of logarithmic functions, always double-check if the chain rule applies.