This calculus worksheet helps students master differentiation using power constant and sum rules through 10 varied practice problems.

To solve these problems, we need to find the derivative of each function. This means we are looking for the rate at which the function changes. We will use three main rules:

1. Power Rule: If you have $x^n$, the derivative is $n \cdot x^{(n-1)}$. (Multiply by the power, then subtract 1 from the power).
2. Constant Rule: The derivative of a constant number (like 5 or -18) is always 0.
3. Sum/Difference Rule: You can take the derivative of each part of the equation separately and add or subtract them.

Let's work through each one step-by-step.

Problem 1: $y = \frac{-17}{x^5}$
* First, rewrite the fraction as a power with a negative exponent: $y = -17x^{-5}$.
* Apply the Power Rule: Multiply the exponent (-5) by the coefficient (-17), then subtract 1 from the exponent.
* $(-5) \cdot (-17) = 85$
* New exponent: $-5 - 1 = -6$
* Result: $y' = 85x^{-6}$ or $\frac{85}{x^6}$

Problem 2: $y = \frac{3}{8}x^{\frac{-3}{13}}$
* Apply the Power Rule: Multiply the exponent ($\frac{-3}{13}$) by the coefficient ($\frac{3}{8}$), then subtract 1 from the exponent.
* Coefficient: $\frac{3}{8} \cdot \frac{-3}{13} = \frac{-9}{104}$
* New exponent: $\frac{-3}{13} - 1 = \frac{-3}{13} - \frac{13}{13} = \frac{-16}{13}$
* Result: $y' = -\frac{9}{104}x^{-\frac{16}{13}}$

Problem 3: $y = -x^3 + 2x^2 + 33x - 90$
* Differentiate each term separately.
* Term 1 ($-x^3$): $3 \cdot (-1)x^{(3-1)} = -3x^2$
* Term 2 ($2x^2$): $2 \cdot 2x^{(2-1)} = 4x$
* Term 3 ($33x$): The power of $x$ is 1. $1 \cdot 33x^{(1-1)} = 33x^0 = 33$.
* Term 4 ($-90$): This is a constant, so the derivative is 0.
* Result: $y' = -3x^2 + 4x + 33$

Problem 4: $y = -18$
* This is just a constant number. There is no variable $x$.
* The derivative of any constant is 0.
* Result: $y' = 0$

Problem 5: $y = -13$
* Similar to problem 4, this is a constant.
* Result: $y' = 0$

Problem 6: $y = -x^2 + 5x^2 + 8x - 48$
* *Tip:* You can simplify the equation first by combining like terms ($-x^2 + 5x^2 = 4x^2$). So, $y = 4x^2 + 8x - 48$.
* Term 1 ($4x^2$): $2 \cdot 4x^{(2-1)} = 8x$
* Term 2 ($8x$): Derivative is 8.
* Term 3 ($-48$): Derivative is 0.
* Result: $y' = 8x + 8$

Problem 7: $y = \frac{6}{7}x^{\frac{4}{7}}$
* Apply the Power Rule: Multiply the exponent ($\frac{4}{7}$) by the coefficient ($\frac{6}{7}$), then subtract 1 from the exponent.
* Coefficient: $\frac{6}{7} \cdot \frac{4}{7} = \frac{24}{49}$
* New exponent: $\frac{4}{7} - 1 = \frac{4}{7} - \frac{7}{7} = -\frac{3}{7}$
* Result: $y' = \frac{24}{49}x^{-\frac{3}{7}}$

Problem 8: $y = \frac{-8}{x^2}$
* Rewrite as a power: $y = -8x^{-2}$.
* Apply the Power Rule: Multiply the exponent (-2) by the coefficient (-8), then subtract 1 from the exponent.
* Coefficient: $(-2) \cdot (-8) = 16$
* New exponent: $-2 - 1 = -3$
* Result: $y' = 16x^{-3}$ or $\frac{16}{x^3}$

Problem 9: $y = \frac{-2}{9}x^{\frac{4}{5}}$
* Apply the Power Rule: Multiply the exponent ($\frac{4}{5}$) by the coefficient ($\frac{-2}{9}$), then subtract 1 from the exponent.
* Coefficient: $\frac{-2}{9} \cdot \frac{4}{5} = \frac{-8}{45}$
* New exponent: $\frac{4}{5} - 1 = \frac{4}{5} - \frac{5}{5} = -\frac{1}{5}$
* Result: $y' = -\frac{8}{45}x^{-\frac{1}{5}}$

Problem 10: $y = -14$
* This is a constant.
* Result: $y' = 0$

Final Answer:
1) $y' = \frac{85}{x^6}$
2) $y' = -\frac{9}{104}x^{-\frac{16}{13}}$
3) $y' = -3x^2 + 4x + 33$
4) $y' = 0$
5) $y' = 0$
6) $y' = 8x + 8$
7) $y' = \frac{24}{49}x^{-\frac{3}{7}}$
8) $y' = \frac{16}{x^3}$
9) $y' = -\frac{8}{45}x^{-\frac{1}{5}}$
10) $y' = 0$