Math Problem Statement

f(x)=ln (3x-tanx)-e^(8x3) + 2^x

Solution

Given the function:

f(x)=ln(3xtanx)e8x3+2xf(x) = \ln(3x - \tan x) - e^{8x^3} + 2^x

Let's break down the structure of the function f(x)f(x):

1. Components of the Function:

  • ln(3xtanx)\ln(3x - \tan x): This logarithmic part requires that the argument 3xtanx3x - \tan x be positive, meaning 3x>tanx3x > \tan x. Also, xx must be in a domain where the tangent function is defined (not at odd multiples of π2\frac{\pi}{2}).

  • e8x3e^{8x^3}: An exponential function that grows very rapidly as xx increases, especially since the exponent is cubic in xx.

  • 2x2^x: Another exponential function, which also increases rapidly, though not as fast as e8x3e^{8x^3} for large values of xx.

2. Domain of f(x)f(x):

  • For ln(3xtanx)\ln(3x - \tan x) to be defined, 3xtanx>03x - \tan x > 0 and x(2n+1)π2x \neq \frac{(2n+1)\pi}{2} for any integer nn because tanx\tan x is undefined at those points.

3. Finding the Derivative:

To find the derivative f(x)f'(x), we use the following rules:

  • The derivative of ln(g(x))\ln(g(x)) is g(x)g(x)\frac{g'(x)}{g(x)}.
  • The derivative of eg(x)e^{g(x)} is eg(x)g(x)e^{g(x)} \cdot g'(x).
  • The derivative of 2x2^x is 2xln22^x \ln 2.

Applying these rules:

[ f'(x) = \frac{d}{dx}[\ln(3x - \tan x)] - \frac{d}{dx}[e^{8x^3}] + \frac{d}{dx}[2^x] ]

Let's compute each part:

  1. ddx[ln(3xtanx)]\frac{d}{dx}[\ln(3x - \tan x)]:

    • Let g(x)=3xtanxg(x) = 3x - \tan x.
    • g(x)=3sec2xg'(x) = 3 - \sec^2 x.
    • So, ddx[ln(3xtanx)]=3sec2x3xtanx\frac{d}{dx}[\ln(3x - \tan x)] = \frac{3 - \sec^2 x}{3x - \tan x}.
  2. ddx[e8x3]\frac{d}{dx}[e^{8x^3}]:

    • Let h(x)=8x3h(x) = 8x^3.
    • h(x)=24x2h'(x) = 24x^2.
    • So, ddx[e8x3]=e8x324x2\frac{d}{dx}[e^{8x^3}] = e^{8x^3} \cdot 24x^2.
  3. ddx[2x]\frac{d}{dx}[2^x]:

    • The derivative of 2x2^x is 2xln22^x \ln 2.

Putting it all together:

f(x)=3sec2x3xtanx24x2e8x3+2xln2f'(x) = \frac{3 - \sec^2 x}{3x - \tan x} - 24x^2 e^{8x^3} + 2^x \ln 2

Would you like a further breakdown of the derivative or an exploration of the properties of this function?

5 Relative Questions to Expand Understanding:

  1. What is the domain of the function f(x)=ln(3xtanx)e8x3+2xf(x) = \ln(3x - \tan x) - e^{8x^3} + 2^x?
  2. How does the presence of exponential terms affect the growth rate of f(x)f(x) as xx increases?
  3. What are the potential points of discontinuity in f(x)f(x) given the logarithmic and tangent functions?
  4. How would you find the critical points of this function?
  5. Can this function have any local maxima or minima given the rapid growth of the exponential components?

Tip:

When dealing with logarithmic functions, always check the domain to ensure the argument of the logarithm is positive and valid throughout the desired range of xx.

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Math Problem Analysis

Mathematical Concepts

Calculus
Logarithmic Functions
Exponential Functions
Tangent Function

Formulas

f(x) = ln(3x - tan x) - e^{8x^3} + 2^x
Derivative of ln(g(x)) is g'(x)/g(x)
Derivative of e^{g(x)} is e^{g(x)} * g'(x)
Derivative of 2^x is 2^x * ln(2)

Theorems

Chain Rule
Logarithmic Differentiation
Exponential Differentiation

Suitable Grade Level

Undergraduate Calculus or Advanced High School Calculus