find the derivative of Y = -0.0000888𝑥3 + -0.00816𝑥2 + 0.4756 + 100

To find the derivative of the given function \( Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100 \), we will differentiate each term with respect to \( x \).

The given function is:

\[
Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100
\]

### Steps to Differentiate:

1. **Differentiate \( -0.0000888x^3 \):**
   \[
   \frac{d}{dx}(-0.0000888x^3) = -0.0000888 \cdot 3x^{3-1} = -0.0002664x^2
   \]

2. **Differentiate \( -0.00816x^2 \):**
   \[
   \frac{d}{dx}(-0.00816x^2) = -0.00816 \cdot 2x^{2-1} = -0.01632x
   \]

3. **Differentiate \( 0.4756x \):**
   \[
   \frac{d}{dx}(0.4756x) = 0.4756
   \]

4. **Differentiate the constant \( 100 \):**
   \[
   \frac{d}{dx}(100) = 0
   \]

### Combining the Results:

Now, we can combine the results from each of these steps:

\[
\frac{dY}{dx} = -0.0002664x^2 - 0.01632x + 0.4756
\]

### Conclusion:

The derivative of the function \( Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100 \) with respect to \( x \) is:

\[
\boxed{\frac{dY}{dx} = -0.0002664x^2 - 0.01632x + 0.4756}
\]

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. What is the second derivative of the function \( Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100 \)?
2. How do you find the critical points of the function using its derivative?
3. What is the significance of the constant term in the original function?
4. How do the coefficients of the polynomial affect the shape of its graph?
5. How would you integrate the function \( Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100 \)?

**Tip:** When differentiating polynomials, remember to apply the power rule \( \frac{d}{dx}(ax^n) = nax^{n-1} \) to each term separately.

Learn how to find the derivative of the polynomial function Y = -0.0000888x^3 - 0.00816x^2 + 0.4756x + 100 with a step-by-step solution. This problem covers differentiation of polynomials using the power rule. Suitable for advanced high school students and learners interested in calculus.

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