PIL import Image import pytesseract

Load the uploaded image

image_path = "/mnt/data/1733605473758retbaokk.jpg" image = Image.open(image_path)

Use OCR to extract text from the image

extracted_text = pytesseract.image_to_string(image) extracted_textThe OCR extraction yielded incomplete or incorrect results. The function given in the image appears to be $f(x) = x^2 (x - 5)^3$. I will proceed to compute its derivative.

Derivative of $f(x) = x^2 (x - 5)^3$

We apply the product rule and chain rule: $f(x) = u \cdot v \quad \text{where } u = x^2 \text{ and } v = (x - 5)^3$

1. Derivative of $u = x^2$: $u' = 2x$

2. Derivative of $v = (x - 5)^3$ using the chain rule: $v' = 3(x - 5)^2 \cdot 1 = 3(x - 5)^2$

Using the product rule: $f'(x) = u'v + uv' = (2x)(x - 5)^3 + (x^2)(3(x - 5)^2)$

Simplify: $f'(x) = 2x(x - 5)^3 + 3x^2(x - 5)^2$

Factor common terms $x(x - 5)^2$: $f'(x) = x(x - 5)^2 \left[ 2(x - 5) + 3x \right]$

Simplify the bracket: $2(x - 5) + 3x = 2x - 10 + 3x = 5x - 10$

Final derivative: $f'(x) = x(x - 5)^2 (5x - 10)$

Would you like me to further analyze or graph this function? Let me know!

Related Questions:

1. How do the product and chain rules apply in other problems?
2. How do we find critical points for $f(x)$ using $f'(x)$?
3. Can this derivative be expanded to standard polynomial form?
4. What are the practical applications of such derivatives?
5. How do you use $f'(x)$ to determine intervals of increase and decrease?

Tip:

Always look for common factors when simplifying derivatives—it makes the expressions more compact and manageable!