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.

For composite functions like $f(x) = \sin(x^2)$, higher-order derivatives use the chain rule multiple times. The first derivative is $f'(x) = \cos(x^2) \cdot 2x$, and the second requires the product rule: $$f''(x) = -\sin(x^2) \cdot 2x \cdot 2x + \cos(x^2) \cdot 2 = 2\cos(x^2) - 4x^2 \sin(x^2)$$, matching choice A. Successive application combines chain and product rules carefully. A common error is stopping after the first chain rule application and ignoring the product in the second derivative. Double-check by expanding terms fully before simplifying. Always identify inner and outer functions to apply rules systematically for any order.

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.

Rational functions' higher derivatives involve increasing negative exponents. The first is y' = $-2/x^3$. The second is $6/x^4$, and third is $-24/x^5$. A common error is stopping at second or sign errors. Constants like 7 vanish early. For inverses, use power rule repeatedly, alternating signs with each differentiation.

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.

Finding m''(x) for m(x) = x²eˣ requires the product rule applied twice. First, m'(x) = 2xeˣ + x²eˣ = (2x + x²)eˣ = (x² + 2x)eˣ. For the second derivative, we apply the product rule to (x² + 2x)eˣ: m''(x) = (2x + 2)eˣ + (x² + 2x)eˣ = [(2x + 2) + (x² + 2x)]eˣ = (x² + 4x + 2)eˣ. A common mistake is losing track of terms when applying the product rule multiple times. When differentiating products involving eˣ, factor out eˣ after each step to simplify the algebra and clearly see the polynomial coefficient pattern.

To find g''(x) for g(x) = sin(2x), we use the chain rule twice. First, g'(x) = cos(2x) · 2 = 2cos(2x), where we multiply by the derivative of the inner function 2x. Applying the chain rule again, g''(x) = -sin(2x) · 2 · 2 = -4sin(2x), since the derivative of cos is -sin and we must again multiply by 2 from the chain rule. A common error is forgetting to apply the chain rule consistently or losing track of the coefficient. When differentiating composite trigonometric functions multiple times, carefully track both the trigonometric derivatives (sin → cos → -sin → -cos) and the chain rule factors from the inner function.

For F(x) = ln(x²), we can use the chain rule or first simplify using logarithm properties: ln(x²) = 2ln|x|. Taking the simpler approach, F'(x) = 2(1/x) = 2/x for x ≠ 0. Then F''(x) = 2(-1/x²) = -2/x². Alternatively, using the chain rule directly: F'(x) = (1/x²)(2x) = 2/x, and F''(x) = -2/x². A common error is forgetting the negative sign when differentiating 1/x or misapplying the chain rule. When dealing with logarithmic functions, simplifying using log properties before differentiating often makes the calculation clearer and reduces errors.

To find p''(x) for p(x) = 1/x² = x⁻², we rewrite using negative exponents and apply the power rule twice. First, p'(x) = -2x⁻³ = -2/x³. Then, p''(x) = -2(-3)x⁻⁴ = 6x⁻⁴ = 6/x⁴. The key is recognizing that differentiating x⁻ⁿ gives -nx⁻ⁿ⁻¹, so negative exponents become more negative with each differentiation. A common error is mishandling the signs or incorrectly applying the power rule to negative exponents. When working with rational functions like 1/xⁿ, converting to negative exponents (x⁻ⁿ) makes differentiation more straightforward and helps avoid quotient rule complications.

To find the second derivative s''(t) of s(t) = t⁴ - 6t² + 3t, we must differentiate twice. First, s'(t) = 4t³ - 12t + 3 using the power rule on each term. Then, differentiating again gives s''(t) = 12t² - 12 + 0 = 12t² - 12. A common error is stopping after the first derivative or incorrectly applying the power rule during the second differentiation. Notice how the constant term 3t becomes 3 after the first derivative, then 0 after the second. When finding higher-order derivatives, systematically apply differentiation rules to each term, tracking how powers decrease with each application.