The task in the image involves approximating derivatives using difference quotients for various functions.

1. Forward difference of $f(x) = \sin(x)$

At $x = -4$ with step size $h = \frac{1}{5} = 0.2$, the forward difference formula is:

$f'(x) \approx \frac{f(x+h) - f(x)}{h}$ Substituting $f(x) = \sin(x)$, $x = -4$, and $h = 0.2$:

$f'(x) \approx \frac{\sin(-4 + 0.2) - \sin(-4)}{0.2}$ This results in an approximation of $-0.72$.

2. Backward difference of $f(x) = x^3 - x + 1$

At $x = -4$ with step size $h = \frac{1}{10} = 0.1$, the backward difference formula is:

$f'(x) \approx \frac{f(x) - f(x-h)}{h}$ Substituting $f(x) = x^3 - x + 1$, $x = -4$, and $h = 0.1$:

$f'(x) \approx \frac{(-4)^3 - (-4) + 1 - \left((-4 - 0.1)^3 - (-4 - 0.1) + 1\right)}{0.1}$ This gives an approximation of $-580.99$.

3. Central difference of $f(x) = \ln(1 + x)$

At $x = 6$ with step size $h = \frac{1}{10} = 0.1$, the central difference formula is:

$f'(x) \approx \frac{f(x+h) - f(x-h)}{2h}$ Substituting $f(x) = \ln(1 + x)$, $x = 6$, and $h = 0.1$:

$f'(x) \approx \frac{\ln(1 + (6 + 0.1)) - \ln(1 + (6 - 0.1))}{2(0.1)}$ This results in an approximation of $0.95$.

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Would you like more details on how these approximations were calculated? Here are 5 related questions:

1. How is the forward difference approximation derived?
2. What are the benefits of using the central difference over forward or backward differences?
3. How does the choice of $h$ affect the accuracy of these approximations?
4. How can you calculate higher-order derivatives using difference methods?
5. What other methods exist for approximating derivatives?

Tip: Smaller step sizes $h$ tend to give more accurate approximations, but very small $h$ can introduce numerical errors due to rounding.