Higher-Order Derivatives (Advanced)

We start by finding the first derivative using the power rule. For f(x) = 4x⁵ - 3x⁴ + 2x² - 7x + 1, we get f'(x) = 20x⁴ - 12x³ + 4x - 7. Next, we differentiate f'(x) to obtain the second derivative. Applying the power rule again gives us f''(x) = 80x³ - 36x² + 4. The constant term -7 from f'(x) has a derivative of zero, so it does not appear in f''(x).

We begin by finding the first derivative using the chain rule. For y = sin(3x), we have dy/dx = cos(3x) · 3 = 3cos(3x). To find the second derivative, we differentiate dy/dx = 3cos(3x) with respect to x. Using the chain rule again, we get d²y/dx² = 3 · (-sin(3x)) · 3 = -9sin(3x). The negative sign comes from the derivative of cosine, and we multiply by 3 again from the chain rule applied to the inner function 3x.

We start by finding the first derivative using the chain rule. For f(x) = e²ˣ, we have f'(x) = e²ˣ · 2 = 2e²ˣ. To find the second derivative, we differentiate f'(x) = 2e²ˣ. Applying the chain rule once more gives us f''(x) = 2 · e²ˣ · 2 = 4e²ˣ. The exponential function e²ˣ maintains its form through differentiation, and we multiply by the constant 2 each time we apply the chain rule.

We begin by finding the first derivative. For g(x) = ln(x²), we can simplify using logarithm properties: g(x) = 2ln(x). Thus, g'(x) = 2 · (1/x) = 2/x. To find the second derivative, we differentiate g'(x) = 2/x, which we can write as g'(x) = 2x^(-1). Using the power rule, we get g''(x) = 2 · (-1) · x^(-2) = -2x^(-2) = -2/x².

We find the derivatives in sequence. Starting with h(x) = x⁴ - 6x³ + 5x, the first derivative is h'(x) = 4x³ - 18x² + 5. Next, we differentiate h'(x) to get the second derivative: h''(x) = 12x² - 36x. Finally, we differentiate h''(x) to obtain the third derivative: h'''(x) = 24x - 36. The constant term from h''(x) has a derivative of zero in each subsequent differentiation, which is why -36 remains as a constant in the third derivative.

We first find the first derivative using the chain rule. For f(x) = (x² + 1)³, we have f'(x) = 3(x² + 1)² · 2x = 6x(x² + 1)². To find f''(x), we use the product rule on f'(x) = 6x(x² + 1)². This gives us f''(x) = 6(x² + 1)² + 6x · 2(x² + 1) · 2x = 6(x² + 1)² + 24x²(x² + 1). Factoring out 6(x² + 1), we get f''(x) = 6(x² + 1)[(x² + 1) + 4x²] = 6(x² + 1)(5x² + 1). Evaluating at x = 1 gives us f''(1) = 6(1 + 1)(5 + 1) = 6 · 2 · 6 = 72.

We recognize that acceleration is the second derivative of position with respect to time. Starting with s(t) = t³ - 9t² + 24t, we find the first derivative (velocity): s'(t) = 3t² - 18t + 24. Next, we find the second derivative (acceleration): s''(t) = 6t - 18. Evaluating at t = 2 gives us s''(2) = 6(2) - 18 = 12 - 18 = -6. The negative value indicates the acceleration is in the opposite direction to the positive direction of motion.

We begin by finding the first derivative using the product rule. For y = x·cos(x), we have dy/dx = cos(x) + x·(-sin(x)) = cos(x) - x·sin(x). To find the second derivative, we differentiate dy/dx using the product rule again on the term x·sin(x). This gives us d²y/dx² = -sin(x) - [sin(x) + x·cos(x)] = -sin(x) - sin(x) - x·cos(x) = -2sin(x) - x·cos(x). Evaluating at x = π/2, we get d²y/dx² = -2sin(π/2) - (π/2)·cos(π/2) = -2(1) - (π/2)(0) = -2.

We find derivatives in sequence. Starting with f(x) = sin(x) + cos(x), the first derivative is f'(x) = cos(x) - sin(x). The second derivative is f''(x) = -sin(x) - cos(x). The third derivative is f'''(x) = -cos(x) + sin(x) = sin(x) - cos(x). Finally, the fourth derivative is f⁽⁴⁾(x) = cos(x) + sin(x) = sin(x) + cos(x). Notice that the fourth derivative returns to the original function, which is a characteristic property of sine and cosine functions.

We start by finding the first derivative using the product rule. For f(x) = x²·eˣ, we have f'(x) = 2x·eˣ + x²·eˣ = (2x + x²)eˣ = (x² + 2x)eˣ. To find the second derivative, we apply the product rule to f'(x) = (x² + 2x)eˣ. This gives us f''(x) = (2x + 2)eˣ + (x² + 2x)eˣ. Combining like terms, we get f''(x) = [(2x + 2) + (x² + 2x)]eˣ = (x² + 4x + 2)eˣ.

We begin by rewriting g(x) = (x² + 1)^(-1). Using the chain rule for the first derivative, we get g'(x) = -1(x² + 1)^(-2) · 2x = -2x/(x² + 1)². For the second derivative, we use the quotient rule on g'(x) = -2x/(x² + 1)². This gives us g''(x) = [(-2)(x² + 1)² - (-2x) · 2(x² + 1) · 2x]/(x² + 1)⁴. Simplifying the numerator, we get -2(x² + 1)² + 8x²(x² + 1) = -2(x² + 1)[(x² + 1) - 4x²] = -2(x² + 1)(1 - 3x²) = 2(x² + 1)(3x² - 1). Therefore, g''(x) = 2(3x² - 1)/(x² + 1)³.

We start by finding the first derivative using the product rule. For h(x) = x·ln(x), we have h'(x) = ln(x) + x·(1/x) = ln(x) + 1. To find the second derivative, we differentiate h'(x) = ln(x) + 1. This gives us h''(x) = 1/x. Evaluating at x = e, we get h''(e) = 1/e.

We begin by finding the first derivative of f(x) = tan(x), which is f'(x) = sec²(x). To find the second derivative, we differentiate f'(x) = sec²(x). Using the chain rule, we have f''(x) = 2sec(x) · sec(x)tan(x) = 2sec²(x)tan(x). Evaluating at x = π/4, we need sec(π/4) = √2 and tan(π/4) = 1. Therefore, f''(π/4) = 2(√2)²(1) = 2(2)(1) = 4.

We start by finding the first derivative of y = arcsin(x), which is dy/dx = 1/√(1 - x²) = (1 - x²)^(-½). To find the second derivative, we use the chain rule on dy/dx = (1 - x²)^(-½). This gives us d²y/dx² = (-½)(1 - x²)^(-3/2) · (-2x) = x(1 - x²)^(-3/2) = x/(1 - x²)^(3/2).

We find the derivatives in sequence using the power rule. Starting with f(x) = x2/3, the first derivative is f'(x) = (⅔)x-1/3 = 2/(3x1/3). The second derivative is f''(x) = (⅔) · (-1/3)x-4/3 = -2/(9x4/3). For the third derivative, we have f'''(x) = (-2/9) · (-4/3)x-7/3 = 8/(27x7/3).