Power Rule: Negative & Fractional Exponents

The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. Here, the exponent n is 8. We multiply the coefficient of x (which is 1) by the exponent 8 to get the new coefficient 8. Then we reduce the exponent by 1, so 8 becomes 7. The result is f'(x) = 8x^(8-1) = 8x⁷. This directly applies the fundamental power rule for derivatives.

For g(x) = 6x⁵, we use both the constant multiple rule and the power rule. First, keep the constant coefficient 6 as is. Then apply the power rule to x⁵: bring down the exponent 5 and reduce the exponent by 1 to get x⁴. Multiply the constant 6 by the new coefficient 5, giving 65 = 30. The final derivative is g'(x) = 30x⁴. The constant multiple rule tells us that the derivative of cf(x) is c*f'(x).

We use the sum rule, which states that the derivative of a sum is the sum of the derivatives. Differentiate each term separately. For x³, the derivative is 3x². For x⁵, the derivative is 5x⁴. Add these results together: f'(x) = 3x² + 5x⁴. The sum rule allows us to handle each term independently before combining the results.

Apply the sum/difference rule and the constant multiple rule to each term separately. For 4x⁷, multiply 4 by the exponent 7 to get 28, and reduce the exponent to 6, giving 28x⁶. For -3x², multiply -3 by the exponent 2 to get -6, and reduce the exponent to 1, giving -6x. Combine these: p'(x) = 28x⁶ - 6x. The constant multiple rule preserves the coefficient in front of each term.

Differentiate each term using the power rule. For 2x⁴: 24 = 8, exponent becomes 3, giving 8x³. For 5x³: 53 = 15, exponent becomes 2, giving 15x². For -x (which is -1x): -1*1 = -1, exponent becomes 0, giving -1. For the constant 7: the derivative is 0. Combine: f'(x) = 8x³ + 15x² - 1. Constants always disappear when taking derivatives.

Using the constant multiple rule and power rule together, the derivative of axⁿ is a times the derivative of xⁿ. The derivative of xⁿ is nxⁿ⁻¹. Therefore, f'(x) = a * nxⁿ⁻¹ = an*xⁿ⁻¹. This shows the general form of the power rule for any monomial.

First, rewrite √(x) as x^(½). Using the power rule, we bring down the exponent 1/2 and multiply it by the coefficient 1, giving 1/2 as the new coefficient. Then we reduce the exponent by 1: (½) - 1 = -1/2. So y' = (½)x^(-½). This is equivalent to 1/(2*√(x)), which is the standard derivative of the square root function.

First find the derivative: f'(x) = 2x - 8. Set this equal to zero: 2x - 8 = 0. Solve for x by adding 8 to both sides: 2x = 8. Then divide by 2: x = 4. This finds the critical point where the tangent line is horizontal, which is important for optimization problems.

The power rule works for negative exponents too. For h(x) = x^(-4), the exponent is -4. We multiply the coefficient (1) by the exponent -4 to get -4 as the new coefficient. Then we subtract 1 from the exponent: -4 - 1 = -5. So h'(x) = -4*x^(-5). The negative exponent means the function is decreasing as x increases, and the negative coefficient in the derivative reflects this decreasing rate.

First combine the terms using exponent addition: x² * x³ = x^(2+3) = x⁵. So g(x) = x⁵. Then apply the power rule: g'(x) = 5x⁴. This problem tests whether you can recognize when to use algebra rules before applying calculus rules.

First expand the expression: (2x³)² = 2² * (x³)² = 4x⁶. So f(x) = 4x⁶. Now apply the power rule: multiply 4 by the exponent 6 to get 24, and reduce the exponent to 5. Therefore f'(x) = 24x⁵. Algebraic simplification is crucial before differentiating.

Velocity is the derivative of position. Find s'(t) by applying the power rule to t³: s'(t) = 3t². Evaluate at t = 2: s'(2) = 3*(2²) = 3*4 = 12. The units are meters per second since position was in meters and time in seconds.

First simplify the expression using exponent rules: x⁸divided by x³ is x^(8-3) = x⁵. So f(x) = x⁵. Now apply the power rule: f'(x) = 5x⁴. The key step is simplifying before differentiating, which makes the problem much easier than using the quotient rule.

Find the first derivative using the power rule: f'(x) = 35x⁴ = 15x⁴. Then differentiate again to get the second derivative. Apply the power rule to 15x⁴: multiply 15 by the exponent 4 to get 60, and reduce the exponent to 3. So f''(x) = 60x³. Taking multiple derivatives requires applying the power rule repeatedly.

First find the derivative function: f'(x) = 6x⁵. Then evaluate at x = 2 by substituting 2 for x. Calculate f'(2) = 6*(2⁵). Since 2⁵ = 32, we have 6*32 = 192. This shows how to find the instantaneous rate of change at a specific point using the power rule.