https://file.aiquickdraw.com/mathbot/file/1726093822576kfr35muu.jpg

The given function is:

\[
y = \frac{4x^3}{3} - x + 2e^x
\]

Let's compute both the first and second derivatives.

### First derivative \( y' \):

We will differentiate each term with respect to \( x \):

1. \(\frac{4x^3}{3}\):
   \[
   \frac{d}{dx}\left(\frac{4x^3}{3}\right) = 4x^2
   \]

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

3. \( 2e^x \):
   \[
   \frac{d}{dx}(2e^x) = 2e^x
   \]

So, the first derivative is:
\[
y' = 4x^2 - 1 + 2e^x
\]

---

### Second derivative \( y'' \):

Now, we differentiate \( y' \) again with respect to \( x \):

1. \( 4x^2 \):
   \[
   \frac{d}{dx}(4x^2) = 8x
   \]

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

3. \( 2e^x \):
   \[
   \frac{d}{dx}(2e^x) = 2e^x
   \]

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

---

### Final Results:

- First derivative: \( y' = 4x^2 - 1 + 2e^x \)
- Second derivative: \( y'' = 8x + 2e^x \)

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!

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

Learn how to compute the first and second derivatives of the function y = (4x^3)/3 - x + 2e^x using differentiation techniques. This solution involves the power rule and the derivative of exponential functions. Suitable for students in undergraduate calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation