Math Problem Statement

Find the first and second derivatives for y = (4x^3)/3 - x + 2e^x

Solution

The given function is:

y=4x33x+2exy = \frac{4x^3}{3} - x + 2e^x

Let's compute both the first and second derivatives.

First derivative yy':

We will differentiate each term with respect to xx:

  1. 4x33\frac{4x^3}{3}: ddx(4x33)=4x2\frac{d}{dx}\left(\frac{4x^3}{3}\right) = 4x^2

  2. x-x: ddx(x)=1\frac{d}{dx}(-x) = -1

  3. 2ex2e^x: ddx(2ex)=2ex\frac{d}{dx}(2e^x) = 2e^x

So, the first derivative is: y=4x21+2exy' = 4x^2 - 1 + 2e^x


Second derivative yy'':

Now, we differentiate yy' again with respect to xx:

  1. 4x24x^2: ddx(4x2)=8x\frac{d}{dx}(4x^2) = 8x

  2. 1-1: ddx(1)=0\frac{d}{dx}(-1) = 0

  3. 2ex2e^x: ddx(2ex)=2ex\frac{d}{dx}(2e^x) = 2e^x

So, the second derivative is: y=8x+2exy'' = 8x + 2e^x


Final Results:

  • First derivative: y=4x21+2exy' = 4x^2 - 1 + 2e^x
  • Second derivative: y=8x+2exy'' = 8x + 2e^x

Let me know if you need more details or have any questions!

Here are 5 related questions for further exploration:

  1. How does the second derivative help determine concavity?
  2. What is the interpretation of the first derivative in terms of slope?
  3. Can you find critical points using the first derivative?
  4. What is the significance of exponential functions in calculus?
  5. How would you solve for inflection points using the second derivative?

Tip: Always simplify the function before taking derivatives to make calculations smoother!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Polynomial Functions
Exponential Functions

Formulas

Derivative of x^n: d/dx(x^n) = n*x^(n-1)
Derivative of e^x: d/dx(e^x) = e^x
Sum rule for differentiation: d/dx(f(x) + g(x)) = f'(x) + g'(x)

Theorems

Power Rule
Exponential Derivative Rule

Suitable Grade Level

Undergraduate Calculus