Binomial roots differentiate with fractional exponents. The first is f'(x) = (1/2)(1 + $x)^{-1/2}$. The second is f''(x) = - (1/4)(1 + $x)^{-3/2}$, reducing exponent. A common error is stopping early or fraction mistakes. Chain rule applies each time. For roots, convert to exponents and differentiate iteratively, updating coefficients.

Product of x and exponential uses product rule. The first is y' = $e^{-x}$(1 - x). The second is y'' = $e^{-x}$(x - 2), factoring properly. A common error is stopping at first or sign errors in exponential. Each differentiation involves both product and chain. For such products, apply rules successively, factoring the exponential out.

For h(x) = e³ˣ, finding the third derivative requires applying the chain rule three times. First, h'(x) = e³ˣ · 3 = 3e³ˣ. Second, h''(x) = 3e³ˣ · 3 = 9e³ˣ. Third, h'''(x) = 9e³ˣ · 3 = 27e³ˣ. Notice the pattern: each differentiation multiplies by 3 (the coefficient in the exponent), so the nth derivative is 3ⁿe³ˣ. A common mistake is forgetting to apply the chain rule at each step or losing track of the accumulating coefficients. For exponential functions with linear exponents like e^(ax), the nth derivative follows the pattern aⁿe^(ax), making higher-order derivatives predictable.

To find s''(x) for s(x) = cos(3x), we apply the chain rule twice. First, s'(x) = -sin(3x)·3 = -3sin(3x). Then s''(x) = -3cos(3x)·3 = -9cos(3x). The pattern emerges: each differentiation introduces a factor of 3 (from the chain rule) and cycles through cos → -sin → -cos → sin. A common mistake is forgetting to multiply by the derivative of the inner function (3) at each step. For trig functions with linear arguments, remember that the nth derivative introduces a factor of the argument's coefficient raised to the nth power.

For u(x) = x²ln x where x > 0, we need the product rule twice. First, u'(x) = 2x ln x + x²(1/x) = 2x ln x + x. For the second derivative, differentiate each term: d/dx(2x ln x) = 2ln x + 2x(1/x) = 2ln x + 2, and d/dx(x) = 1. Therefore, u''(x) = 2ln x + 2 + 1 = 2ln x + 3. A common error is forgetting that d/dx(ln x) = 1/x when applying the product rule to the first term. Always carefully track each application of differentiation rules when products involve logarithms.

Tangent's derivatives involve secant and tangent terms. The first is f'(x) = $sec^2$(x). The second is f''(x) = 2 $sec^2$(x) tan(x), using chain rule. A common stopping error is only computing the first. Express in trig identities if simplifying. A transferable strategy for trig derivatives is to use known identities and chain rule repeatedly.

To find the second derivative f''(x) of f(x) = 3x⁴ - 5x² + 7x, we need to differentiate twice. First, f'(x) = 12x³ - 10x + 7 using the power rule. Then, differentiating again gives f''(x) = 36x² - 10. A common error is stopping after the first derivative or forgetting to differentiate the constant term (which becomes 0). Notice how the degree decreases by 1 with each differentiation. For polynomial functions, systematically apply the power rule to each term, reducing the exponent by 1 and multiplying by the original exponent each time.

For q(x) = e^(2x), we need the third derivative q'''(x). Using the chain rule repeatedly: q'(x) = 2e^(2x), q''(x) = 4e^(2x), and q'''(x) = 8e^(2x). Each differentiation multiplies by the derivative of the exponent (which is 2), giving us powers of 2: 2¹, 2², 2³. A common mistake is forgetting to apply the chain rule consistently or losing track of the coefficient. For exponential functions with linear exponents, the nth derivative of e^(kx) is $k^n$·e^(kx).

Logarithmic derivatives simplify, but higher orders need care. The first is f'(x) = 1/x. The second is f''(x) = $-1/x^2$, with negative sign. A common error is stopping at first or ignoring the constant in ln(5x). Domain x > 0. For logs, differentiate successively, noting inverse square patterns.

Binomial powers differentiate with chain rule and signs. The first is f'(x) = -4(1 - $x)^3$. The second is f''(x) = 12(1 - $x)^2$, flipping positive. A common stopping error is only first derivative. Note the sign changes from the inner derivative. A transferable strategy is to apply chain rule multiple times, tracking the inner function's derivative.