This problem involves differentiating arccos of a cubic function. The derivative of arccos(u) is -1/√(1-u²) times u'. Here, u = x³, so u' = 3x². Applying the chain rule: t'(x) = -1/√(1-(x³)²) · 3x² = -3x²/√(1-x⁶). Choice A omits the negative sign essential to arccos derivatives. The negative sign distinguishes arccos from arcsin derivatives and must always be included. Remember: arcsin and arctan derivatives are positive, while arccos and arccot derivatives are negative.

This requires differentiating arccos, which has the derivative formula -1/√(1-u²) times u'. Here, u = 2x+3, so u' = 2. Applying the chain rule: h'(x) = -1/√(1-(2x+3)²) · 2 = -2/√(1-(2x+3)²). Choice A omits the negative sign that is essential to the arccos derivative formula. The negative sign in arccos derivatives reflects that cosine decreases as its angle increases in the range [0,π]. Always include the negative sign when differentiating arccos.

This problem requires differentiating an inverse trigonometric function, specifically arctangent of a composite function. The derivative of arctan(u) is 1/(1+u²) times the derivative of u. Here, u = 3x², so u' = 6x. Applying the chain rule: f'(x) = 1/(1+(3x²)²) · 6x = 6x/(1+9x⁴). A common error would be to use the arcsine derivative formula 1/√(1-u²), which gives the incorrect radical form in choice A. When differentiating inverse trig functions, always match the correct derivative formula to the specific inverse function.

This problem requires differentiating arccos with a linear composite function. The derivative of arccos(u) is -1/√(1-u²), noting the negative sign that distinguishes it from arcsin. Applying the chain rule with u = 5x: h'(x) = -1/√(1-(5x)²) · d/dx(5x) = -1/√(1-(5x)²) · 5 = -5/√(1-25x²). Choice A incorrectly omits the negative sign, giving 5/√(1-(5x)²), which would be the derivative of arcsin(5x) instead. When differentiating inverse trig functions, remember that arccos is the only one of the three main inverse trig derivatives with a negative sign.

This problem involves differentiating arccos with a linear composite function. The derivative of arccos(u) is -1/√(1-u²), and we apply the chain rule with u = 1-4s. Differentiating: r'(s) = -1/√(1-(1-4s)²) · d/ds(1-4s) = -1/√(1-(1-4s)²) · (-4) = 4/√(1-(1-4s)²). The two negative signs cancel to give a positive result. Choice B incorrectly keeps the negative sign as -4/√(1-(1-4s)²). For inverse trig derivatives, carefully track signs when both arccos (negative derivative) and decreasing inner functions are involved.

This problem involves differentiating the inverse tangent function, a key skill in AP Calculus BC. The derivative formula for arctan(u) is u' divided by 1 plus u squared. Here, u equals 5x minus 1, so u' is 5. Applying the formula gives 5 over 1 plus (5x minus 1) squared. A tempting distractor is choice A, which uses the arcsin formula with a square root instead of the correct arctan denominator. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.

This problem involves differentiating the inverse sine function, a key skill in AP Calculus BC. The derivative formula for arcsin(u) is u' divided by the square root of 1 minus u squared. Here, u equals x over 2, so u' is 1 over 2. Applying the formula gives (1 over 2) over the square root of 1 minus (x over 2) squared, which is 1 over (2 times the square root of 1 minus (x over 2) squared). A tempting distractor is choice A, which forgets the chain rule factor of 1 over 2 from u'. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.

This problem involves differentiating the inverse tangent function, a key skill in AP Calculus BC. The derivative formula for $arctan(u)$ is $\frac{u'}{1 + u^2}$. Here, $u = \frac{2}{x}$, so $u' = -\frac{2}{x^2}$. Applying the formula gives $-\frac{2}{x^2}$ divided by $1 + \left(\frac{2}{x}\right)^2$. A tempting distractor is choice C, which omits the negative sign from $u'$. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.

This problem involves differentiating the inverse cosine function, a key skill in AP Calculus BC. The derivative formula for $arccos(u)$ is negative $u'$ divided by the square root of $1 - u^2$. Here, u equals 2x, so $u'$ is 2. Applying the formula gives negative 2 over the square root of $1 - (2x)^2$, which is negative 2 over the square root of $1 - 4x^2$. A tempting distractor is choice A, which omits the negative sign inherent to the arccos derivative. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.

This problem involves differentiating the inverse sine function, a key skill in AP Calculus BC. The derivative formula for arcsin(u) is u' divided by the square root of 1 minus u squared. Here, u equals 3x squared, so u' is 6x. Applying the formula gives 6x over the square root of 1 minus (3x squared) squared, which simplifies to 6x over the square root of 1 minus 9x to the fourth. A tempting distractor is choice C, which includes a negative sign, but that would be incorrect because the arcsin derivative formula does not inherently include a negative unlike arccos. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.