Power Rule in Applied Problems

The power rule states that for y = xⁿ, the derivative is n*xⁿ⁻¹. Here n = 7. Multiply the coefficient (1) by the exponent 7 to get 7. Then reduce the exponent by 1: 7 - 1 = 6. So dy/dx = 7x⁶. This is the fundamental application of the power rule to a simple monomial.

Apply the constant multiple rule first, keeping the coefficient 8. Then apply the power rule to x³: bring down the exponent 3 and reduce it to 2, giving 3x². Multiply the constant 8 by this result: 8*3x² = 24x². So f'(x) = 24x². The constant multiple rule preserves the coefficient throughout the differentiation.

For negative exponents, the power rule works the same way. For g(x) = x^(-7), multiply the coefficient (1) by the exponent -7 to get -7. Then subtract 1 from the exponent: -7 - 1 = -8. So g'(x) = -7x^(-8). The negative exponent becomes more negative after differentiation, and the negative coefficient appears because we multiply by the negative exponent.

The power rule applies to fractional exponents as well. For h(x) = x^(5/2), multiply the coefficient (1) by the exponent 5/2 to get 5/2. Then reduce the exponent by 1: (5/2) - 1 = (5/2) - (2/2) = 3/2. So h'(x) = (5/2)x^(3/2). This shows the derivative of a radical-like function.

Apply the difference rule, which says the derivative of a difference is the difference of derivatives. Differentiate each term separately. For x^9: 9x^8. For x⁴: 4x³, but since it's subtracted, we get -4x³. Combine: f'(x) = 9x⁸- 4x³. Each term is handled independently using the power rule.

Use the sum rule to differentiate each term separately. For 3x⁶: multiply 3 by the exponent 6 to get 18, and reduce exponent to 5, giving 18x⁵. For 2x⁴: multiply 2 by the exponent 4 to get 8, and reduce exponent to 3, giving 8x³. Combine: p'(x) = 18x⁵ + 8x³. The sum rule lets us handle each term as a separate problem.

The constant coefficient is 1/4. Apply the power rule to x^8: the derivative is 8x⁷. Multiply this by the constant 1/4: (1/4)*8x⁷ = 2x⁷. So f'(x) = 2x⁷. When the coefficient is a fraction, the multiplication step simplifies the coefficient.

First find the derivative function: f'(x) = 5x⁴. Now evaluate at x = -1. Calculate f'(-1) = 5*(-1)⁴. Since (-1)⁴ = 1 (any negative number to an even power is positive), we have 5*1 = 5. So f'(-1) = 5. The negative input becomes positive because of the even exponent in the derivative.

First find the first derivative: f'(x) = 54x³ = 20x³. Now differentiate again to get the second derivative. Apply the power rule to 20x³: multiply 20 by the exponent 3 to get 60, and reduce the exponent to 2, giving 60x². So f''(x) = 60x². Each differentiation reduces the exponent by one and adjusts the coefficient.

First simplify by dividing each term in the numerator by x³. Then x¹⁰ divided by x³ equals x^(10-3) = x⁷. And x⁷ divided by x³ equals x^(7-3) = x⁴. So f(x) = x⁷ + x⁴. Now apply the power rule to each term: the derivative of x⁷ is 7x⁶, and the derivative of x⁴ is 4x³. So f'(x) = 7x⁶ + 4x³. Simplifying before differentiating avoids using the quotient rule.

First expand the binomial: (x + 3)² = x² + 6x + 9. Now differentiate term by term. The derivative of x² is 2x. The derivative of 6x is 6. The derivative of the constant 9 is 0. So g'(x) = 2x + 6. This can be factored as 2(x + 3), showing a pattern. Expanding before differentiating is key for polynomial forms.

First rewrite √(x⁷) as x^(7/2). Now apply the power rule: multiply the coefficient (1) by the exponent 7/2 to get 7/2. Then reduce the exponent by 1: (7/2) - 1 = (7/2) - (2/2) = 5/2. So dy/dx = (7/2)x^(5/2). This is equivalent to (7/2)*√(x⁵). The exponent 7/2 represents the seventh power of x under a square root.

First find the derivative: f'(x) = 2x + 3. Set this equal to 5: 2x + 3 = 5. Subtract 3 from both sides: 2x = 2. Divide by 2: x = 1. So at x = 1, the instantaneous rate of change of the function is 5. This demonstrates how to use the derivative to find specific points with given slope values.

Marginal cost is the derivative of the cost function. Find C'(x) by differentiating: C'(x) = 32x + 20 = 6x + 20. Now evaluate at x = 10: C'(10) = 6*10 + 20 = 60 + 20 = 80. So the marginal cost at production level 10 is 80 dollars per additional item. This shows how derivatives model real-world rates of change in economics.

Since x = 1 is in the region where x ≤ 2, we use the first piece of the function: f(x) = x². The derivative of x² is 2x. Evaluate this at x = 1: f'(1) = 2*1 = 2. When finding derivatives of piecewise functions, we only consider the piece that contains the point of interest. The other piece is irrelevant for this calculation.