Math Problem Statement

Compute the derivatives of the following functions: y = log2(2x^3 + 3x - 4), f(t) = ln(ln(t)), g(x) = (log7(x^2 + x + 1))^5.

Solution

The image shows three problems involving derivatives and logarithms. Let's solve them one by one.

Problem 1:

Given: y=log2(2x3+3x4)y = \log_2(2x^3 + 3x - 4)

To differentiate yy with respect to xx, we will use the chain rule. Recall the derivative of a logarithmic function: ddxloga(u)=1uln(a)dudx\frac{d}{dx} \log_a(u) = \frac{1}{u \ln(a)} \cdot \frac{du}{dx} where u=2x3+3x4u = 2x^3 + 3x - 4 and a=2a = 2.

First, let's find dudx\frac{du}{dx}: ddx(2x3+3x4)=6x2+3\frac{d}{dx} (2x^3 + 3x - 4) = 6x^2 + 3

Now, using the formula for the derivative of a logarithm: dydx=1(2x3+3x4)ln(2)(6x2+3)\frac{dy}{dx} = \frac{1}{(2x^3 + 3x - 4) \ln(2)} \cdot (6x^2 + 3)

Thus, the derivative is: dydx=6x2+3(2x3+3x4)ln(2)\frac{dy}{dx} = \frac{6x^2 + 3}{(2x^3 + 3x - 4) \ln(2)}


Problem 2:

Given: f(t)=ln(ln(t))f(t) = \ln(\ln(t))

We apply the chain rule again. The derivative of ln(u)\ln(u) is 1u\frac{1}{u}, and we have u=ln(t)u = \ln(t).

First, differentiate the outer function: ddtln(ln(t))=1ln(t)\frac{d}{dt} \ln(\ln(t)) = \frac{1}{\ln(t)}

Now, multiply by the derivative of the inner function ln(t)\ln(t): ddtln(t)=1t\frac{d}{dt} \ln(t) = \frac{1}{t}

Thus, the derivative is: f(t)=1tln(t)f'(t) = \frac{1}{t \ln(t)}


Problem 3:

Given: g(x)=(log7(x2+x+1))5g(x) = \left( \log_7(x^2 + x + 1) \right)^5

This is a composite function, so we apply the chain rule. First, we differentiate the outer function u5u^5 with respect to u=log7(x2+x+1)u = \log_7(x^2 + x + 1), then differentiate the inner function.

First, the outer function: dduu5=5u4\frac{d}{du} u^5 = 5u^4

Now, apply the chain rule: g(x)=5(log7(x2+x+1))4ddxlog7(x2+x+1)g'(x) = 5 \left( \log_7(x^2 + x + 1) \right)^4 \cdot \frac{d}{dx} \log_7(x^2 + x + 1)

We now differentiate the logarithm log7(x2+x+1)\log_7(x^2 + x + 1): ddxloga(u)=1uln(a)dudx\frac{d}{dx} \log_a(u) = \frac{1}{u \ln(a)} \cdot \frac{du}{dx} where u=x2+x+1u = x^2 + x + 1 and a=7a = 7.

First, compute dudx\frac{du}{dx}: ddx(x2+x+1)=2x+1\frac{d}{dx} (x^2 + x + 1) = 2x + 1

Thus: ddxlog7(x2+x+1)=2x+1(x2+x+1)ln(7)\frac{d}{dx} \log_7(x^2 + x + 1) = \frac{2x + 1}{(x^2 + x + 1) \ln(7)}

Now substitute back into the expression for g(x)g'(x): g(x)=5(log7(x2+x+1))42x+1(x2+x+1)ln(7)g'(x) = 5 \left( \log_7(x^2 + x + 1) \right)^4 \cdot \frac{2x + 1}{(x^2 + x + 1) \ln(7)}


Do you need further clarification on any of these solutions? Let me know if you'd like a detailed breakdown of any steps!

Here are five related questions you might want to explore:

  1. How do you differentiate logb(f(x))\log_b(f(x)) when f(x)f(x) is more complex?
  2. Can you apply the product rule and chain rule together in logarithmic functions?
  3. How would you handle logarithmic derivatives when the base is a variable rather than a constant?
  4. What are the applications of logarithmic differentiation in real-world problems?
  5. How would the rules change if natural logarithms were involved?

Tip: Remember that logarithmic differentiation is especially useful when dealing with products or quotients of functions, particularly when those functions are raised to powers.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Logarithmic Differentiation
Chain Rule
Logarithmic Functions
Derivatives

Formulas

Derivative of log base a: d/dx[log_a(u)] = 1 / (u ln(a)) * du/dx
Derivative of natural logarithm: d/dx[ln(u)] = 1 / u * du/dx
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)

Theorems

Chain Rule
Logarithmic Differentiation Theorem

Suitable Grade Level

College-Level Calculus or Advanced High School