How to solve for the derivative given this information: #f(x+h)-f(x) = 2 h x^2 + 3 h x - 8 h^2 x + 8 h^2 - 4 h^3# ?

Question

How to solve for the derivative given this information: #f(x+h)-f(x) = 2 h x^2  + 3 h x  - 8 h^2 x  + 8 h^2  - 4 h^3# ?

Ezekiel Alexander · Accepted Answer

Given: #f(x+h)-f(x) = 2 h x^2 + 3 h x - 8 h^2 x + 8 h^2 - 4 h^3#

Divide both sides of the equation by h:

#(f(x+h)-f(x))/h = 2x^2 + 3x - 8 h x + 8 h - 4 h^2#

Compute the limit as #h to 0#

#lim_(h to 0) (f(x+h)-f(x))/h = 2x^2 + 3x#

This is the definition of the derivative.

Luke Alvarez · Answer

undefined To solve for the derivative using the provided information $f(x+h)-f(x) = 2hx^2 + 3hx - 8hx^2 + 8h^2 - 4h^3$:

1. Start with the definition of the difference quotient:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

2. Substitute the given expression for $f(x+h)-f(x)$ into the difference quotient:
$$f'(x) = \lim_{h \to 0} \frac{2hx^2 + 3hx - 8hx^2 + 8h^2 - 4h^3}{h}$$

3. Simplify the expression by canceling out common terms and factoring out $h$:
$$f'(x) = \lim_{h \to 0} \frac{(2x^2 - 8x^2)h + (3x)h + (8h^2 - 4h^3)}{h}$$
$$f'(x) = \lim_{h \to 0} \frac{(-6x^2 + 3x)h + (8h^2 - 4h^3)}{h}$$
$$f'(x) = \lim_{h \to 0} (-6x^2 + 3x) + \frac{(8h^2 - 4h^3)}{h}$$

4. Evaluate the limit as $h$ approaches 0:
$$f'(x) = -6x^2 + 3x + \lim_{h \to 0} (8h - 4h^2)$$

5. The limit term $\lim_{h \to 0} (8h - 4h^2)$ becomes 0 when $h$ tends to 0.

6. Therefore, the derivative $f'(x)$ simplifies to:
$$f'(x) = -6x^2 + 3x$$