Math Problem Statement

Differentiate the functions, find derivatives, and determine the domain for various logarithmic functions as shown in the math problems.

Solution

Let's work through each of the problems you've uploaded:

Problem 6: Differentiate the function F(s)=ln(ln(2s))F(s) = \ln(\ln(2s))

  1. Outer derivative (ln): The derivative of ln(u)\ln(u) with respect to uu is 1u\frac{1}{u}.

    • Here, u=ln(2s)u = \ln(2s), so the first part of the derivative is 1ln(2s)\frac{1}{\ln(2s)}.
  2. Inner derivative (ln(2s)): The derivative of ln(2s)\ln(2s) is 12s\frac{1}{2s}, using the chain rule.

Thus, the derivative of F(s)F(s) is: F(s)=1ln(2s)12s=12sln(2s)F'(s) = \frac{1}{\ln(2s)} \cdot \frac{1}{2s} = \frac{1}{2s \ln(2s)}

Problem 7: Find yy' and yy'' for y=x2ln(5x)y = x^2 \ln(5x)

  1. First derivative yy':

    • This is a product rule problem. The product rule is (uv)=uv+uv(uv)' = u'v + uv'.
    • Let u=x2u = x^2 and v=ln(5x)v = \ln(5x).
    • Derivative of u=x2u = x^2 is u=2xu' = 2x.
    • Derivative of v=ln(5x)v = \ln(5x) is v=1xv' = \frac{1}{x} (since ln(5x)=ln(5)+ln(x)\ln(5x) = \ln(5) + \ln(x)).

    Applying the product rule: y=2xln(5x)+x21x=2xln(5x)+xy' = 2x \ln(5x) + x^2 \cdot \frac{1}{x} = 2x \ln(5x) + x

  2. Second derivative yy'':

    • Use the product rule again for 2xln(5x)2x \ln(5x) and combine with the derivative of xx.

    The second derivative is: y=2ln(5x)+2x1x+1=2ln(5x)+2+1=2ln(5x)+3y'' = 2 \ln(5x) + 2x \cdot \frac{1}{x} + 1 = 2 \ln(5x) + 2 + 1 = 2 \ln(5x) + 3

Problem 8: Differentiate f(x)=x1ln(x7)f(x) = \frac{x}{1 - \ln(x - 7)} and find the domain.

  1. Quotient rule for derivative: The quotient rule is (uv)=uvuvv2\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.

    • u=xu = x, so u=1u' = 1.
    • v=1ln(x7)v = 1 - \ln(x - 7), so v=1x7v' = -\frac{1}{x - 7} (derivative of ln(x7)\ln(x - 7)).

    Applying the quotient rule:

    = \frac{1 - \ln(x - 7) + \frac{x}{x - 7}}{(1 - \ln(x - 7))^2}$$
  2. Domain:

    • The domain is restricted by the natural logarithm, so x7>0x - 7 > 0, meaning x>7x > 7.
    • Additionally, 1ln(x7)01 - \ln(x - 7) \neq 0, so ln(x7)1\ln(x - 7) \neq 1, or x7ex - 7 \neq e, which gives x7+ex \neq 7 + e.

    Therefore, the domain is (7,7+e)(7+e,)(7, 7 + e) \cup (7 + e, \infty).

Problem 9: Use logarithmic differentiation to find the derivative of y=x8cos(x)y = x^8 \cos(x)

  1. Take the natural logarithm of both sides: ln(y)=ln(x8cos(x))=ln(x8)+ln(cos(x))=8ln(x)+ln(cos(x))\ln(y) = \ln(x^8 \cos(x)) = \ln(x^8) + \ln(\cos(x)) = 8 \ln(x) + \ln(\cos(x))

  2. Differentiate implicitly: ddx[ln(y)]=1yy=81x+sin(x)cos(x)=81xtan(x)\frac{d}{dx} [\ln(y)] = \frac{1}{y} \cdot y' = 8 \cdot \frac{1}{x} + \frac{-\sin(x)}{\cos(x)} = 8 \cdot \frac{1}{x} - \tan(x)

    Therefore, multiplying both sides by y=x8cos(x)y = x^8 \cos(x), we get: y=x8cos(x)(8xtan(x))y' = x^8 \cos(x) \left( \frac{8}{x} - \tan(x) \right)


Let me know if you'd like more details or have any questions! Here are five related questions:

  1. How is the chain rule applied to logarithmic functions?
  2. Can you find the second derivative of F(s)=ln(ln(2s))F(s) = \ln(\ln(2s))?
  3. What are the conditions for applying the quotient rule?
  4. Can you derive the product rule for a more complex function, like y=xnexy = x^n e^x?
  5. How does logarithmic differentiation simplify the process for differentiating products of functions?

Tip: Always double-check the domain of logarithmic functions to ensure the argument is positive!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Logarithmic Differentiation
Quotient Rule
Product Rule
Chain Rule
Implicit Differentiation

Formulas

Derivative of ln(u): d/dx [ln(u)] = 1/u * du/dx
Product Rule: (uv)' = u'v + uv'
Quotient Rule: (u/v)' = (u'v - uv') / v^2
Derivative of ln(x): d/dx [ln(x)] = 1/x
Implicit Differentiation

Theorems

Chain Rule
Product Rule
Quotient Rule
Logarithmic Differentiation

Suitable Grade Level

Grade 12, College-level Calculus