Math Problem Statement

derivative of f(x) = (ln(x - 1) - 1)/(ln(x - 1))^2

Solution

To find the derivative of the function:

f(x)=ln(x1)1(ln(x1))2f(x) = \frac{\ln(x - 1) - 1}{(\ln(x - 1))^2}

we will use the quotient rule. The quotient rule states that for a function of the form f(x)=u(x)v(x)f(x) = \frac{u(x)}{v(x)}, the derivative f(x)f'(x) is given by:

f(x)=u(x)v(x)u(x)v(x)(v(x))2.f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}.

Step 1: Define u(x)u(x) and v(x)v(x)

  • u(x)=ln(x1)1u(x) = \ln(x - 1) - 1
  • v(x)=(ln(x1))2v(x) = (\ln(x - 1))^2

Step 2: Compute u(x)u'(x) and v(x)v'(x)

Derivative of u(x)u(x)

u(x)=ln(x1)1u(x) = \ln(x - 1) - 1

Using the chain rule:

u(x)=1x1.u'(x) = \frac{1}{x - 1}.

Derivative of v(x)v(x)

v(x)=(ln(x1))2v(x) = (\ln(x - 1))^2

Using the chain rule and the power rule:

v(x)=2ln(x1)1x1=2ln(x1)x1.v'(x) = 2 \ln(x - 1) \cdot \frac{1}{x - 1} = \frac{2 \ln(x - 1)}{x - 1}.

Step 3: Apply the Quotient Rule

Substituting u(x)u(x), v(x)v(x), u(x)u'(x), and v(x)v'(x) into the quotient rule:

f(x)=u(x)v(x)u(x)v(x)(v(x))2.f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}.

Plugging in the expressions:

f(x)=(1x1)(ln(x1))2(ln(x1)1)(2ln(x1)x1)(ln(x1)2)2.f'(x) = \frac{\left(\frac{1}{x - 1}\right)(\ln(x - 1))^2 - (\ln(x - 1) - 1)\left(\frac{2 \ln(x - 1)}{x - 1}\right)}{(\ln(x - 1)^2)^2}.

Step 4: Simplify the Numerator

Factor out 1x1\frac{1}{x - 1} from both terms in the numerator:

f(x)=1x1[(ln(x1))22ln(x1)(ln(x1)1)](ln(x1))4.f'(x) = \frac{\frac{1}{x - 1} \left[ (\ln(x - 1))^2 - 2\ln(x - 1)(\ln(x - 1) - 1) \right]}{(\ln(x - 1))^4}.

Simplify inside the brackets:

(ln(x1))22ln(x1)(ln(x1)1)=(ln(x1))22(ln(x1))2+2ln(x1).(\ln(x - 1))^2 - 2\ln(x - 1)(\ln(x - 1) - 1) = (\ln(x - 1))^2 - 2(\ln(x - 1))^2 + 2\ln(x - 1).

Combine like terms:

=(ln(x1))2+2ln(x1).= -(\ln(x - 1))^2 + 2\ln(x - 1).

Thus, the expression for f(x)f'(x) becomes:

f(x)=1x1[(ln(x1))2+2ln(x1)](ln(x1))4.f'(x) = \frac{\frac{1}{x - 1} \left[ -(\ln(x - 1))^2 + 2\ln(x - 1) \right]}{(\ln(x - 1))^4}.

Step 5: Final Expression

Combining the terms:

f(x)=(ln(x1))2+2ln(x1)(x1)(ln(x1))4.f'(x) = \frac{-(\ln(x - 1))^2 + 2\ln(x - 1)}{(x - 1)(\ln(x - 1))^4}.

This is the derivative of the function f(x)f(x).


Do you need a more detailed explanation or simplification?

Related Questions:

  1. What is the derivative of g(x)=ln(x)(ln(x))2g(x) = \frac{\ln(x)}{(\ln(x))^2}?
  2. How do you apply the quotient rule to f(x)=exx2f(x) = \frac{e^x}{x^2}?
  3. What is the second derivative of f(x)=ln(x1)1(ln(x1))2f(x) = \frac{\ln(x - 1) - 1}{(\ln(x - 1))^2}?
  4. How do you simplify expressions involving logarithmic derivatives?
  5. What are the critical points of f(x)=ln(x1)1(ln(x1))2f(x) = \frac{\ln(x - 1) - 1}{(\ln(x - 1))^2}?

Tip:

When dealing with derivatives of logarithmic functions, always double-check if the chain rule applies.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Quotient Rule
Logarithmic Functions

Formulas

Quotient Rule: f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2
Derivative of ln(x): d/dx[ln(x)] = 1/x
Power Rule: d/dx[x^n] = n * x^(n-1)

Theorems

Quotient Rule in Calculus
Chain Rule for Derivatives

Suitable Grade Level

Undergraduate Calculus or Advanced High School Calculus (Grades 11-12)