Higher-Order Derivatives Fundamentals

First, find the first derivative using the power rule: f'(x) = 4x³. Then differentiate again using the power rule: the derivative of 4x³ is 4 · 3x² = 12x². So the second derivative is 12x².

First, find the first derivative: dy/dx = 15x² - 4x + 7. Then differentiate again: the derivative of 15x² is 30x, the derivative of -4x is -4, and the derivative of 7 is 0. So d²y/dx² = 30x - 4.

The first derivative of sin(x) is cos(x). The second derivative is -sin(x). Which when differentiated again, gives -cos(x).

The derivative of e2x is 2e2x (using the chain rule). The second derivative is again 2 · 2e2x = 4e2x. The third derivative is 2 · 4e2x = 8e2x. Each differentiation multiplies by another factor of 2.

First derivative: f'(x) = 15x⁴ - 12x² + 2. Second derivative: derivative of 15x⁴ is 60x³, derivative of -12x² is -24x, derivative of 2 is 0. So f''(x) = 60x³ - 24x.

All three notations are standard ways to represent the second derivative of a function y = f(x). f''(x) is the prime notation, d²y/dx² is the Leibniz notation, and y'' is the shorthand version of the prime notation when the function is named y.

To find the second derivative, differentiate the first derivative once more. The derivative of 6x² is 12x, the derivative of -8x is -8, and the derivative of 3 is 0. So f''(x) = 12x - 8.

First derivative: 6x⁵. Second: 30x⁴. Third: 120x³. Fourth: 360x².

The first derivative f'(x) gives the slope of the function. The second derivative f''(x) tells how that slope is changing, which determines whether the graph is concave up (f'' > 0) or concave down (f''

First derivative: f'(x) = -3sin(3x) (chain rule: derivative of cos(u) is -sin(u) times derivative of inside). Second derivative: derivative of -3sin(3x) is -3 · cos(3x) · 3 = -9cos(3x).

First derivative: 5x⁴. Second: 20x³. Third: 60x².

The first derivative of position s(t) is velocity v(t). The second derivative of position (or first derivative of velocity) is acceleration a(t). So s''(t) = a(t).

Let u = 2x + 1, so f(x) = u³. First derivative: f'(x) = 3u² · 2 = 6(2x + 1)². Second derivative: derivative of 6(2x + 1)² is 6 · 2(2x + 1) · 2 = 24(2x + 1).

The fourth derivative can be written as f^(4)(x) or d⁴y/dx⁴ or y^{(4)}. The notation f^(4)(x) with the superscript in parentheses is standard for fourth and higher-order derivatives to avoid confusion with prime marks.

Work backwards by differentiating twice and matching. If f(x) = 2x³ - 3x² + ax + b, then the first derivative f'(x) = 6x² - 6x + a, and the second derivative f''(x) = 12x - 6, which matches exactly. The linear term ax + b disappears when differentiated twice, so any linear function can be added without changing the second derivative.