Answer: Since $\cos(y) = \sqrt{1-x^2}$, we get $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$

Use the Pythagorean identity: From $\sin(y)=x$ and $\sin^2(y)+\cos^2(y)=1$, we get $\cos^2(y)=1-x^2$. Since $y \in [-\pi/2, \pi/2]$, $\cos(y) \geq 0$, so $\cos(y) = \sqrt{1-x^2}$. Why the correct answer works: Option C correctly identifies $\cos(y) = \sqrt{1-x^2}$, yielding $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$. Why distractors fail: Option A uses $1+x^2$ instead of $1-x^2$. Option B omits the square root. Option D incorrectly claims $\cos(y) = x$, confusing it with $\sin(y)=x$.

Answer: It is correct.

Apply the formula with chain rule: For $u = x^2$ with $x > 1$ so $u > 1$: $\frac{d}{dx}[\text{arcsec}(u)] = \frac{u'}{|u|\sqrt{u^2-1}} = \frac{2x}{x^2\sqrt{x^4-1}}$ (since $u = x^2 > 0$, $|u| = x^2$). Why the correct answer works: The student's work is correct. Option A is the right assessment. Why distractors fail: Option B removes the $x^2$ denominator, which comes from the $|u|$ factor. Option C omits the chain rule factor $u' = 2x$. Option D incorrectly adds a negative sign; arcsec's derivative is positive for $u > 1$.

Answer: $\frac{2\arcsin(x)}{\sqrt{1-x^2}}$

Apply the chain rule to the power: Let $u = \arcsin(x)$, so $f(x) = u^2$. Then $f'(x) = 2u \cdot u' = 2\arcsin(x) \cdot \frac{1}{\sqrt{1-x^2}}$. Why the correct answer works: Option C correctly gives $\frac{2\arcsin(x)}{\sqrt{1-x^2}}$. Why distractors fail: Option A drops the $\arcsin(x)$ factor from the power rule. Option B has $1-x^2$ instead of $\sqrt{1-x^2}$ in the denominator. Option D uses the exponent 2 on $\arcsin(x)$ instead of applying the power rule correctly.

Answer: $y = \frac{1}{2}(x - 1) + \frac{\pi}{4}$

Find the point and slope: At $x=1$: $y = \arctan(1) = \frac{\pi}{4}$. The slope is $y' = \frac{1}{1+x^2}\big|_{x=1} = \frac{1}{2}$. Write the tangent line: Using point-slope form: $y - \frac{\pi}{4} = \frac{1}{2}(x - 1)$, or $y = \frac{1}{2}(x-1) + \frac{\pi}{4}$. This is Option A. Why distractors fail: Option B uses a slope of 1 instead of $\frac{1}{2}$. Option C uses $y$-value 1 instead of $\frac{\pi}{4}$. Option D uses slope $\frac{1}{4}$ instead of $\frac{1}{2}$.

Answer: $e^x \arccos(x) - \frac{e^x}{\sqrt{1-x^2}}$

Apply the product rule: With $f = e^x$ and $g = \arccos(x)$: $\frac{d}{dx}[fg] = e^x \arccos(x) + e^x \cdot \left(\frac{-1}{\sqrt{1-x^2}}\right) = e^x \arccos(x) - \frac{e^x}{\sqrt{1-x^2}}$. Why the correct answer works: Option A correctly applies the product rule with the negative sign from the arccos derivative. Why distractors fail: Option B uses a plus sign instead of minus for the second term, forgetting the negative in the arccos derivative. Option C only includes the second term. Option D omits the $e^x$ factor in the second term.

Answer: $\arctan(x) + \frac{x}{1+x^2}$

Apply the product rule: Using $\frac{d}{dx}[f \cdot g] = f'g + fg'$ with $f = x$ and $g = \arctan(x)$: the derivative is $1 \cdot \arctan(x) + x \cdot \frac{1}{1+x^2}$. Why the correct answer works: Option A correctly combines both terms from the product rule: $\arctan(x) + \frac{x}{1+x^2}$. Why distractors fail: Option B only includes the second term, omitting $\arctan(x)$. Option C writes $\frac{1}{1+x^2}$ instead of $\frac{x}{1+x^2}$ for the second term. Option D only applies part of the product rule without the $\arctan(x)$ term.

Answer: $\frac{2x}{\sqrt{1-x^4}}$

Apply the chain rule: Let $u = x^2$, so $\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}} = \frac{2x}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}}$. Why the correct answer works: Option C correctly includes both the chain rule factor $2x$ and the correct expression $1-x^4$ under the radical. Why distractors fail: Option A omits the chain rule factor $2x$. Option B incorrectly uses $1-x^2$ instead of $1-(x^2)^2 = 1-x^4$. Option D uses $x$ instead of $2x$ as the chain rule factor.

Answer: $\frac{1}{\sqrt{1-x^2}}$

Recall the standard formula: The derivative of $\arcsin(x)$ is a standard result: $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$, valid for $|x| < 1$. Why the correct answer works: Option B is $\frac{1}{\sqrt{1-x^2}}$, which matches the standard derivative formula exactly. Why distractors fail: Option A, $\frac{1}{1+x^2}$, is the derivative of $\arctan(x)$. Option C is the derivative of $\arccos(x)$ (note the negative sign). Option D, $\frac{1}{x\sqrt{x^2-1}}$, is the derivative of $\text{arcsec}(x)$.

Answer: Both students are correct; $1-(2x+1)^2 = -4x^2-4x$ and the derivative is $\frac{2}{\sqrt{-4x^2-4x}}$.

Verify the derivative: With $u = 2x+1$ and $u' = 2$: $f'(x) = \frac{2}{\sqrt{1-(2x+1)^2}}$. This is correct. Verify the simplification: $1-(2x+1)^2 = 1 - (4x^2+4x+1) = -4x^2-4x$. The second student's algebra is correct. Why distractors fail: Option A incorrectly claims the chain rule factor is $\frac{1}{2}$; since $\frac{d}{dx}(2x+1)=2$, the factor is 2. Option C expands $1-(2x+1)^2$ incorrectly. Option D is wrong because the derivative of $\arcsin$ is always positive (the sign depends on the chain rule factor, which is positive here).

Answer: The student forgot to apply the chain rule; the correct answer is $\frac{5}{1+25x^2}$.

Identify the mistake: The student correctly used $\frac{1}{1+u^2}$ with $u=5x$, yielding $\frac{1}{1+25x^2}$, but forgot to multiply by $u' = 5$. Why the correct answer works: Option B correctly identifies the missing chain rule factor. The full derivative is $\frac{5}{1+25x^2}$. Why distractors fail: Option A is wrong because the student did use the correct base formula for arctan. Option C incorrectly introduces a negative sign; arctan's derivative is positive. Option D is wrong because squaring the 5 to get 25 is correct when substituting $u=5x$ into $1+u^2$.

Answer: $\frac{1}{x(1+(\ln x)^2)}$

Apply the chain rule: Let $u = \ln(x)$. Then $\frac{d}{dx}[\arctan(u)] = \frac{u'}{1+u^2} = \frac{1/x}{1+(\ln x)^2} = \frac{1}{x(1+(\ln x)^2)}$. Why the correct answer works: Option A correctly applies the chain rule with $u' = 1/x$. Why distractors fail: Option B forgets the chain rule factor $1/x$. Option C incorrectly places $\ln x$ in the numerator. Option D incorrectly uses $x^2$ instead of $x$ in the denominator, as if the chain rule factor were $1/x^2$ rather than $1/x$.

Answer: $\frac{1}{1+x^2}$

Recall the formula: The standard derivative is $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$. Why the correct answer works: Option C matches the standard formula exactly. Why distractors fail: Option A is the derivative of $\arcsin(x)$. Option B has an incorrect negative sign (that would correspond to $\text{arccot}(x)$). Option D, $\sec^2(x)$, is the derivative of $\tan(x)$, not $\arctan(x)$.

Answer: $\frac{3}{1+9x^2}$

Apply the chain rule: Using the chain rule: $\frac{d}{dx}[\arctan(u)] = \frac{u'}{1+u^2}$ where $u=3x$ and $u'=3$. Compute the result: Substituting: $\frac{3}{1+(3x)^2} = \frac{3}{1+9x^2}$. This matches Option B. Why distractors fail: Option A forgets the factor of $3$ from the chain rule. Option C fails to square the $3$ inside the parentheses, writing $3x^2$ instead of $9x^2$. Option D inverts the chain rule factor.

Answer: $\frac{2}{(1+x^2)^2}$

Differentiate each term: The derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$. For $\frac{x}{1+x^2}$, apply the quotient rule: $\frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2}$. Combine the terms: $g'(x) = \frac{1}{1+x^2} + \frac{1-x^2}{(1+x^2)^2} = \frac{1+x^2}{(1+x^2)^2} + \frac{1-x^2}{(1+x^2)^2} = \frac{2}{(1+x^2)^2}$. Why distractors fail: Option A has the wrong denominator (not squared). Option C results from an error in the quotient rule. Option D introduces a factor of $(1-x^2)$ that doesn't survive after combining fractions correctly.

Answer: $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$

Identify the underlying identity: Since $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$ for all $x \in [-1,1]$, differentiating both sides gives $\frac{d}{dx}[\arcsin(x)] + \frac{d}{dx}[\arccos(x)] = 0$. Why the correct answer works: Option B states the identity whose differentiation directly produces the given equation. Why distractors fail: Option A is the Pythagorean identity for trig functions, not inverse trig. Option C is a valid identity for arctan and arccot but does not explain the arcsin/arccos relationship. Option D is a useful identity but does not directly explain why the sum of derivatives is zero.

Answer: $\frac{1}{\arcsin(x) \cdot \sqrt{1-x^2}}$

Apply the chain rule twice: Using $\frac{d}{dx}[\ln(u)] = \frac{u'}{u}$ with $u = \arcsin(x)$, we get $\frac{1}{\arcsin(x)} \cdot \frac{1}{\sqrt{1-x^2}} = \frac{1}{\arcsin(x)\sqrt{1-x^2}}$. Why the correct answer works: Option A correctly chains the derivative of $\ln$ with the derivative of $\arcsin(x)$. Why distractors fail: Option B only differentiates the $\ln$ layer and forgets the derivative of $\arcsin(x)$. Option C has $\sqrt{1-x^2}$ in the numerator instead of the denominator. Option D replaces $\sqrt{1-x^2}$ with $x$ incorrectly.

Answer: $\frac{-e^x}{\sqrt{1-e^{2x}}}$

Apply the chain rule to arccos: Using $\frac{d}{dx}[\arccos(u)] = \frac{-u'}{\sqrt{1-u^2}}$ with $u = e^x$ and $u' = e^x$: the result is $\frac{-e^x}{\sqrt{1-(e^x)^2}} = \frac{-e^x}{\sqrt{1-e^{2x}}}$. Why the correct answer works: Option C includes both the negative sign from the arccos derivative and the chain rule factor $e^x$. Why distractors fail: Option A is missing the negative sign. Option B forgets the chain rule factor $e^x$. Option D has $e^{2x}-1$ instead of $1-e^{2x}$ under the radical.

Answer: $\frac{-1}{\sqrt{1-x^2}}$

Recall the derivative of arccos: The standard result is $\frac{d}{dx}[\arccos(x)] = \frac{-1}{\sqrt{1-x^2}}$. Why the correct answer works: Option B gives $\frac{-1}{\sqrt{1-x^2}}$, which is correct. Note the negative sign distinguishes it from the derivative of $\arcsin(x)$. Why distractors fail: Option A is the derivative of $\arcsin(x)$ (missing the negative). Option C is the derivative of $\text{arccot}(x)$, not $\arccos(x)$. Option D has $x^2-1$ instead of $1-x^2$ and lacks the negative sign.

Answer: $\frac{1}{|x|\sqrt{x^2-1}}$

Recall the arcsec derivative: The derivative of $\text{arcsec}(x)$ is $\frac{1}{|x|\sqrt{x^2-1}}$, defined for $|x|>1$. Why the correct answer works: Option A matches the standard formula. The absolute value ensures the derivative is positive on the standard domain. Why distractors fail: Option B has $1-x^2$ under the radical (which would be negative for $|x|>1$) and lacks the absolute value. Option C has an incorrect negative sign (that pattern matches arccsc). Option D is the derivative of $\arctan(x)$.