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).

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, 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.

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.

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.

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 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.

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.

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.

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.

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.