Inverse Trig Derivatives - Calculus 1

1. What is an equation of the line tangent to the graph of $f(x) = \arcsec(2x)$ at the point where $x = \frac{1}{\sqrt{3}}$?
   Answer: $y = 3x - \sqrt{3} + \frac{\pi}{6}$ When finding the equation of a tangent line to any function, you need two key pieces: a point on the curve and the slope at that point. For inverse trigonometric functions like arcsecant, this means carefully applying the derivative formula and evaluating at the given x-value. First, find the y-coordinate when $x = \frac{1}{\sqrt{3}}$. Since $f(x) = \arcsec(2x)$, we have $f\left(\frac{1}{\sqrt{3}}\right) = \arcsec\left(\frac{2}{\sqrt{3}}\right) = \arcsec\left(\frac{2\sqrt{3}}{3}\right) = \frac{\pi}{6}$. So our point is $\left(\frac{1}{\sqrt{3}}, \frac{\pi}{6}\right)$. Next, find the slope using the derivative. The derivative of $\arcsec(u)$ is $\frac{1}{|u|\sqrt{u^2-1}}$, so $f'(x) = \frac{2}{|2x|\sqrt{(2x)^2-1}} = \frac{1}{|x|\sqrt{4x^2-1}}$. At $x = \frac{1}{\sqrt{3}}$: $f'\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\frac{1}{\sqrt{3}}\sqrt{\frac{4}{3}-1}} = \frac{\sqrt{3}}{\sqrt{\frac{1}{3}}} = 3$. Using point-slope form: $y - \frac{\pi}{6} = 3\left(x - \frac{1}{\sqrt{3}}\right)$, which simplifies to $y = 3x - \sqrt{3} + \frac{\pi}{6}$. This matches choice C. Choice A has $\frac{\pi}{3}$ instead of $\frac{\pi}{6}$, likely from incorrectly evaluating the arcsecant. Choice B has the wrong slope and completely different form. Choice D omits the $-\sqrt{3}$ term that comes from distributing the slope through the point-slope formula. Remember: with inverse trig derivatives, double-check your algebra carefully—the expressions involve nested radicals and fractions that are easy to miscompute.
2. If $y = \frac{\arcsin(x)}{x}$, find the value of $y'(\frac{1}{\sqrt{2}})$.
   Answer: $2 - \frac{\pi}{2}$ When you encounter a function that's a quotient involving inverse trigonometric functions, you'll need the quotient rule combined with the derivative of inverse trig functions. Here, $y = \frac{\arcsin(x)}{x}$ requires careful application of both. Using the quotient rule $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$, where $f(x) = \arcsin(x)$ and $g(x) = x$: - $f'(x) = \frac{1}{\sqrt{1-x^2}}$ (derivative of arcsine) - $g'(x) = 1$ So $y' = \frac{\frac{1}{\sqrt{1-x^2}} \cdot x - \arcsin(x) \cdot 1}{x^2} = \frac{\frac{x}{\sqrt{1-x^2}} - \arcsin(x)}{x^2}$ Now substitute $x = \frac{1}{\sqrt{2}}$: First, $\arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$ (this is a special angle). Next, $\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1-\frac{1}{2}} = \frac{1}{\sqrt{2}}$ Therefore: $y'\left(\frac{1}{\sqrt{2}}\right) = \frac{\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} - \frac{\pi}{4}}{\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{1 - \frac{\pi}{4}}{\frac{1}{2}} = 2\left(1 - \frac{\pi}{4}\right) = 2 - \frac{\pi}{2}$ Choice A gives $\sqrt{2} - \frac{\pi}{2}$, which incorrectly leaves the first term unsimplified. Choice C gives $1 - \frac{\pi}{4}$, forgetting to divide by $x^2 = \frac{1}{2}$ (missing the factor of 2). Choice D gives $2 - \pi$, which has the wrong coefficient for $\pi$. Remember: when evaluating derivatives at specific points involving inverse trig functions, know your special angles ($\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$) and be methodical with the quotient rule—algebraic errors are common here.
3. For the curve defined by the equation $y + \arctan(xy) = 1$, find the derivative $\frac{dy}{dx}$.
   Answer: $-\frac{y}{1 + x + x^2y^2}$ This problem requires implicit differentiation. We differentiate both sides of the equation with respect to $x$. $\frac{d}{dx}(y + \arctan(xy)) = \frac{d}{dx}(1)$ $\frac{dy}{dx} + \frac{d}{dx}(\arctan(xy)) = 0$ For the $\arctan(xy)$ term, we use the chain rule. Let $u=xy$. The derivative is $\frac{1}{1+u^2} \cdot \frac{du}{dx}$. We need the product rule to find $\frac{du}{dx}$: $\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$. So, the equation becomes: $\frac{dy}{dx} + \frac{1}{1+(xy)^2} \left(y + x\frac{dy}{dx}\right) = 0$ $\frac{dy}{dx} + \frac{y}{1+x^2y^2} + \frac{x}{1+x^2y^2}\frac{dy}{dx} = 0$ Now, we solve for $\frac{dy}{dx}$: $\frac{dy}{dx} \left(1 + \frac{x}{1+x^2y^2}\right) = -\frac{y}{1+x^2y^2}$ Multiply the entire equation by $(1+x^2y^2)$ to clear the denominator: $\frac{dy}{dx} (1+x^2y^2 + x) = -y$ $\frac{dy}{dx} = -\frac{y}{1+x+x^2y^2}$ - Distractor B results from an error in the product rule when differentiating $xy$, specifically treating $x$ as a constant. - Distractor C is a sign error in the final step of solving for $\frac{dy}{dx}$. - Distractor D represents a significant algebraic error when isolating the $\frac{dy}{dx}$ term.
4. The function $f(x) = \arcsin(3x-1)$ is defined on the closed interval $[0, \frac{2}{3}]$. On what open interval is its derivative, $f'(x)$, defined?
   Answer: $(0, \frac{2}{3})$ When you encounter questions about the domain of a derivative, you need to consider where the original function is both defined AND differentiable. For inverse trigonometric functions like arcsine, this requires careful attention to domain restrictions. First, let's find $f'(x)$. Using the chain rule with $\frac{d}{dx}[\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$: $f'(x) = \frac{1}{\sqrt{1-(3x-1)^2}} \cdot 3 = \frac{3}{\sqrt{1-(3x-1)^2}}$ The derivative exists only where the denominator is non-zero and real, meaning $1-(3x-1)^2 > 0$. This gives us $(3x-1)^2 < 1$, so $|3x-1| < 1$, which means $-1 < 3x-1 < 1$. Solving: $0 < 3x < 2$, so $0 < x < \frac{2}{3}$. However, we must also consider that $f(x)$ is only defined on $[0, \frac{2}{3}]$. The derivative cannot exist at the endpoints of a closed interval in the usual sense, and our calculation confirms the derivative is undefined at $x = 0$ and $x = \frac{2}{3}$ (where the denominator equals zero). Answer A $(-\infty, \infty)$ ignores domain restrictions entirely. Answer B $[0, \frac{2}{3}]$ incorrectly includes endpoints where the derivative is undefined. Answer C $(-\frac{1}{3}, \frac{1}{3})$ uses the general domain restriction but ignores that $f(x)$ itself is only defined on $[0, \frac{2}{3}]$. Answer D $(0, \frac{2}{3})$ correctly combines both the derivative's existence condition and the original function's domain. Study tip: For inverse trig derivatives, always check where the expression under the square root is positive, then intersect with the original function's domain.
5. What is the derivative of $f(x) = \arcsin(\cos(x))$ for $x \in (0, \pi)$?
   Answer: $-1$ When you encounter a composite function involving inverse trigonometric functions, you need to apply the chain rule carefully and pay attention to the domain restrictions. To find the derivative of $f(x) = \arcsin(\cos(x))$, apply the chain rule. The derivative of $\arcsin(u)$ is $\frac{1}{\sqrt{1-u^2}}$, so: $f'(x) = \frac{1}{\sqrt{1-\cos^2(x)}} \cdot \frac{d}{dx}[\cos(x)] = \frac{1}{\sqrt{\sin^2(x)}} \cdot (-\sin(x)) = \frac{-\sin(x)}{|\sin(x)|}$ Since we're given $x \in (0, \pi)$, and $\sin(x) > 0$ throughout this interval, we have $|\sin(x)| = \sin(x)$. Therefore: $f'(x) = \frac{-\sin(x)}{\sin(x)} = -1$ Answer choice D is correct. Answer A, $\frac{1}{|\sin(x)|}$, represents what you'd get if you forgot the negative sign from the derivative of cosine and missed simplifying the fraction. Answer B, $1$, is the result if you made both the sign error and forgot that the derivative of the outer function $\arcsin$ isn't simply 1. Answer C, $-\cot(x)$, might appear if you incorrectly applied trigonometric identities or confused the derivatives of inverse trig functions. Study tip: When differentiating composite functions with inverse trig functions, always check whether absolute value expressions can be simplified using the given domain. The domain often determines the sign, leading to much simpler final answers than the initial calculation suggests.
6. If $h(t) = \frac{\arctan(t)}{1+t^2}$, what is $h'(1)$?
   Answer: $\frac{2-\pi}{8}$ When you encounter a function that's a quotient involving trigonometric or inverse trigonometric functions, you'll need the quotient rule combined with knowledge of derivative formulas. To find $h'(t)$ where $h(t) = \frac{\arctan(t)}{1+t^2}$, apply the quotient rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ Here, $f(t) = \arctan(t)$ and $g(t) = 1+t^2$. You need: - $f'(t) = \frac{1}{1+t^2}$ (derivative of arctangent) - $g'(t) = 2t$ Applying the quotient rule: $h'(t) = \frac{\frac{1}{1+t^2} \cdot (1+t^2) - \arctan(t) \cdot 2t}{(1+t^2)^2} = \frac{1 - 2t\arctan(t)}{(1+t^2)^2}$ At $t = 1$: $h'(1) = \frac{1 - 2(1)\arctan(1)}{(1+1)^2} = \frac{1 - 2 \cdot \frac{\pi}{4}}{4} = \frac{1 - \frac{\pi}{2}}{4} = \frac{2-\pi}{8}$ This confirms answer D is correct. Answer A ($-\frac{1}{4}$) likely comes from forgetting the arctangent term entirely. Answer B ($\frac{1-\pi/2}{2}$) makes the right calculation but uses the wrong denominator—forgetting to square $(1+t^2)$ in the quotient rule. Answer C ($\frac{1-\pi}{4}$) incorrectly uses $\arctan(1) = \frac{\pi}{2}$ instead of $\frac{\pi}{4}$, a common arctangent value mistake. Remember: $\arctan(1) = \frac{\pi}{4}$, and always square the denominator when applying the quotient rule.
7. Let $f(x) = \tan(\frac{x}{2})$ for $x \in (-\pi, \pi)$. Let $g(x)$ be the inverse function of $f(x)$. What is the value of $g'(1)$?
   Answer: $1$ When you encounter a question about the derivative of an inverse function, remember the inverse function derivative rule: if $g(x)$ is the inverse of $f(x)$, then $g'(a) = \frac{1}{f'(g(a))}$. To find $g'(1)$, you need two things: the value $g(1)$ and the derivative $f'(x)$. Since $g(x)$ is the inverse of $f(x) = \tan(\frac{x}{2})$, we have $g(1) = x$ where $f(x) = 1$. This means $\tan(\frac{x}{2}) = 1$, so $\frac{x}{2} = \frac{\pi}{4}$, giving us $x = \frac{\pi}{2}$. Therefore, $g(1) = \frac{\pi}{2}$. Next, find $f'(x)$. Using the chain rule: $f'(x) = \sec^2(\frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2}\sec^2(\frac{x}{2})$. Now apply the formula: $g'(1) = \frac{1}{f'(g(1))} = \frac{1}{f'(\frac{\pi}{2})} = \frac{1}{\frac{1}{2}\sec^2(\frac{\pi}{4})}$. Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, we have $\sec^2(\frac{\pi}{4}) = 2$. Therefore: $g'(1) = \frac{1}{\frac{1}{2} \cdot 2} = \frac{1}{1} = 1$. Choice (A) $\frac{1}{4}$ likely comes from incorrectly squaring the $\frac{1}{2}$ factor. Choice (B) $\frac{1}{2}$ results from forgetting to account for $\sec^2(\frac{\pi}{4}) = 2$. Choice (C) $2$ happens if you flip the inverse derivative formula. Remember: always identify what value the inverse function takes at your point of interest before applying the inverse derivative rule.
8. At what value of $x$ is the tangent line to the graph of $y = \arctan(x) - \ln(\sqrt{1+x^2})$ horizontal?
   Answer: $x=1$ When you encounter a question asking where a tangent line is horizontal, you're looking for points where the derivative equals zero. A horizontal tangent line has slope zero, so you need to find where $\frac{dy}{dx} = 0$. Let's find the derivative of $y = \arctan(x) - \ln(\sqrt{1+x^2})$. First, rewrite the logarithm: $\ln(\sqrt{1+x^2}) = \ln((1+x^2)^{1/2}) = \frac{1}{2}\ln(1+x^2)$. So $y = \arctan(x) - \frac{1}{2}\ln(1+x^2)$. Taking the derivative: $\frac{dy}{dx} = \frac{1}{1+x^2} - \frac{1}{2} \cdot \frac{2x}{1+x^2} = \frac{1}{1+x^2} - \frac{x}{1+x^2} = \frac{1-x}{1+x^2}$ Setting this equal to zero: $\frac{1-x}{1+x^2} = 0$. Since the denominator is always positive, we need the numerator to equal zero: $1-x = 0$, which gives us $x = 1$. Let's check each answer choice: - **A) $x=0$**: At $x=0$, $\frac{dy}{dx} = \frac{1-0}{1+0^2} = 1 \neq 0$ - **B) $x=-1$**: At $x=-1$, $\frac{dy}{dx} = \frac{1-(-1)}{1+(-1)^2} = \frac{2}{2} = 1 \neq 0$ - **C) $x=1$**: At $x=1$, $\frac{dy}{dx} = \frac{1-1}{1+1^2} = 0$ ✓ - D) This is incorrect since we found $x=1$ works. Remember: horizontal tangent lines always occur where the derivative equals zero (and exists). Set up the equation $f'(x) = 0$ and solve algebraically.
9. If $g(x) = x \arcsin(3x)$, then $g'(x)$ is:
   Answer: $\arcsin(3x) + \frac{3x}{\sqrt{1-9x^2}}$ This problem requires the product rule and the chain rule. The product rule is $(uv)' = u'v + uv'$. Here, let $u=x$ and $v=\arcsin(3x)$. First, $u' = 1$. Second, to find $v'$, we use the chain rule on $\arcsin(3x)$. The derivative of $\arcsin(w)$ is $\frac{1}{\sqrt{1-w^2}}$. Let $w=3x$, so $w'=3$. Thus, $v' = \frac{1}{\sqrt{1-(3x)^2}} \cdot 3 = \frac{3}{\sqrt{1-9x^2}}$. Applying the product rule: $g'(x) = (1) \cdot \arcsin(3x) + x \cdot \left(\frac{3}{\sqrt{1-9x^2}}\right) = \arcsin(3x) + \frac{3x}{\sqrt{1-9x^2}}$. - Distractor B results from forgetting the chain rule when differentiating $\arcsin(3x)$, missing the factor of 3. - Distractor C is the result of only differentiating the $\arcsin(3x)$ part and multiplying by $x$, but completely ignoring the product rule structure. - Distractor D incorrectly uses the derivative formula for $\arctan(w)$ instead of $\arcsin(w)$.
10. The angle $\theta$ of a searchlight, in radians, is given by $\theta(t) = \operatorname{arccot}(\frac{t}{100})$, where $t$ is time in seconds. Find the rate of change of the angle in radians per second when $t=100$.
    Answer: $-\frac{1}{200}$ This problem tests your ability to differentiate inverse trigonometric functions, specifically the arccotangent function. When you see a rate of change question involving inverse trig functions, you need to apply the chain rule along with the derivative formula for that specific function. To find the rate of change, you need to differentiate $\theta(t) = \operatorname{arccot}\left(\frac{t}{100}\right)$ with respect to time. The derivative of $\operatorname{arccot}(u)$ is $-\frac{1}{1+u^2} \cdot \frac{du}{dt}$. Here, $u = \frac{t}{100}$, so $\frac{du}{dt} = \frac{1}{100}$. Applying the chain rule: $\frac{d\theta}{dt} = -\frac{1}{1+\left(\frac{t}{100}\right)^2} \cdot \frac{1}{100}$ When $t = 100$: $\frac{d\theta}{dt} = -\frac{1}{1+\left(\frac{100}{100}\right)^2} \cdot \frac{1}{100} = -\frac{1}{1+1} \cdot \frac{1}{100} = -\frac{1}{2} \cdot \frac{1}{100} = -\frac{1}{200}$ Answer A ($-\frac{1}{2}$) occurs if you forget to apply the chain rule and miss the $\frac{1}{100}$ factor. Answer B ($\frac{1}{200}$) results from forgetting the negative sign in the arccotangent derivative formula. Answer D ($-\frac{1}{100}$) happens if you incorrectly evaluate the denominator as just 1 instead of $1+1=2$. Remember that inverse trig derivatives often involve negative signs and reciprocals. Always double-check the derivative formula and carefully apply the chain rule when the input isn't simply $x$.
11. Evaluate the limit: $\lim_{h\to 0} \frac{\arctan(1+h) - \frac{\pi}{4}}{h}$
    Answer: $\frac{1}{2}$ When you see a limit of the form $\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$, you're looking at the definition of a derivative. This limit represents $f'(a)$, where $f(x) = \arctan(x)$ and $a = 1$. Since $\arctan(1) = \frac{\pi}{4}$, our limit becomes $\lim_{h\to 0} \frac{\arctan(1+h) - \arctan(1)}{h}$, which is exactly the derivative of $\arctan(x)$ at $x = 1$. The derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$. At $x = 1$, this gives us $\frac{1}{1+1^2} = \frac{1}{2}$. Therefore, the answer is B) $\frac{1}{2}$. Let's see why the other options are wrong. Choice A) $\frac{\pi}{4}$ might tempt you because that's the value of $\arctan(1)$, but remember we're finding the derivative (rate of change), not the function value itself. Choice C) $1$ could come from forgetting the derivative formula and thinking the slope of $\arctan(x)$ is always 1. Choice D) $2$ might result from incorrectly computing $1 + 1^2$ as the numerator instead of the denominator in the derivative formula. The key strategy here is recognizing the derivative definition pattern. Whenever you see $\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$, immediately think "this is $f'(a)$" and find the derivative of the function, then evaluate it at the given point.
12. For the function $f(x) = \arctan(e^x - x)$, at which value of $x$ is the slope of the tangent line equal to zero?
    Answer: $x=0$ When you're asked to find where the slope of a tangent line equals zero, you're looking for critical points where the derivative is zero. This requires finding $f'(x)$ and solving $f'(x) = 0$. To find the derivative of $f(x) = \arctan(e^x - x)$, you'll need the chain rule. The derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \cdot u'$, where $u = e^x - x$. First, find $u' = \frac{d}{dx}(e^x - x) = e^x - 1$. Therefore: $f'(x) = \frac{1}{1+(e^x-x)^2} \cdot (e^x - 1)$ For the slope to equal zero, we need $f'(x) = 0$. Since the fraction $\frac{1}{1+(e^x-x)^2}$ is always positive (the denominator is always greater than 1), the only way for $f'(x) = 0$ is if the numerator's factor $(e^x - 1) = 0$. Solving $e^x - 1 = 0$ gives us $e^x = 1$, so $x = 0$. This confirms answer B is correct. Looking at the other options: A) At $x = 1$, we have $e^1 - 1 = e - 1 \neq 0$, so the derivative isn't zero. C) At $x = -1$, we get $e^{-1} - 1 = \frac{1}{e} - 1 \neq 0$, so this also fails. D) is incorrect because we've found that such a value does exist. Study tip: When finding critical points of composite functions, remember that a fraction equals zero only when its numerator is zero (and denominator is non-zero). Focus on setting the variable factors in the numerator equal to zero.
13. If $y = \arctan(\sqrt{x^2 - 1})$ for $x > 1$, what is $\frac{dy}{dx}$?
    Answer: $\frac{1}{x\sqrt{x^2 - 1}}$ This requires the chain rule. Let $u = \sqrt{x^2 - 1}$. Then $y = \arctan(u)$. The chain rule states $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. We have $\frac{dy}{du} = \frac{1}{1+u^2}$ and $\frac{du}{dx} = \frac{1}{2\sqrt{x^2-1}}(2x) = \frac{x}{\sqrt{x^2-1}}$. Substituting $u = \sqrt{x^2-1}$ into $\frac{dy}{du}$ gives $\frac{1}{1+(\sqrt{x^2-1})^2} = \frac{1}{1+x^2-1} = \frac{1}{x^2}$. Therefore, $\frac{dy}{dx} = \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2-1}} = \frac{1}{x\sqrt{x^2-1}}$.
14. Let $f(x) = \arccsc(\ln x)$. What is the value of $f'(e^2)$?
    Answer: $-\frac{1}{2e^2\sqrt{3}}$ When you encounter the composition of inverse trigonometric functions with other functions, you need to apply the chain rule carefully and know the derivatives of inverse trig functions. To find $f'(x)$ where $f(x) = \arccsc(\ln x)$, use the chain rule. The derivative of $\arccsc(u)$ is $\frac{-1}{|u|\sqrt{u^2-1}}$, and here $u = \ln x$. So $f'(x) = \frac{-1}{|\ln x|\sqrt{(\ln x)^2-1}} \cdot \frac{d}{dx}[\ln x] = \frac{-1}{|\ln x|\sqrt{(\ln x)^2-1}} \cdot \frac{1}{x}$ At $x = e^2$, we have $\ln(e^2) = 2$. Since $2 > 0$, we have $|\ln(e^2)| = 2$. Therefore: $f'(e^2) = \frac{-1}{2\sqrt{2^2-1}} \cdot \frac{1}{e^2} = \frac{-1}{2\sqrt{3}} \cdot \frac{1}{e^2} = \frac{-1}{2e^2\sqrt{3}}$ This confirms answer (C). (A) gives the positive version, which misses the negative sign in the arccosecant derivative formula. (B) is incorrect because the derivative exists everywhere in the domain of $\arccsc(\ln x)$, which requires $|\ln x| \geq 1$. Since $\ln(e^2) = 2 > 1$, the derivative exists. (D) correctly has the negative sign but omits the factor of $\frac{1}{e^2}$ that comes from the chain rule. Study tip: Always memorize that inverse trig derivatives often have negative signs (arccsc, arccot, arcsec), and remember to apply the chain rule when the input isn't just $x$.
15. Let $y = \arctan(x)$. What is the value of $\frac{d^2y}{dx^2}$ at $x=1$?
    Answer: $-\frac{1}{2}$ When you encounter a problem asking for the second derivative of an inverse trigonometric function, you need to find the first derivative, then differentiate again using appropriate rules. Starting with $y = \arctan(x)$, let's find the first derivative. The derivative of $\arctan(x)$ is $\frac{dy}{dx} = \frac{1}{1+x^2}$. To find the second derivative, we differentiate $\frac{1}{1+x^2}$ using the chain rule. We can rewrite this as $(1+x^2)^{-1}$, so: $\frac{d^2y}{dx^2} = \frac{d}{dx}[(1+x^2)^{-1}] = -1 \cdot (1+x^2)^{-2} \cdot 2x = \frac{-2x}{(1+x^2)^2}$ At $x = 1$: $\frac{d^2y}{dx^2} = \frac{-2(1)}{(1+1^2)^2} = \frac{-2}{4} = -\frac{1}{2}$ Answer choice A) gives $-1$, which you might get if you forgot to apply the chain rule correctly when differentiating $(1+x^2)^{-1}$ and missed the factor of $2x$. Answer choice B) gives $\frac{1}{2}$, which is the correct magnitude but wrong sign—this happens if you drop the negative sign from the power rule. Answer choice C) gives $1$, which might result from confusing this with the first derivative evaluated at $x = 0$. Remember that second derivatives of inverse trig functions often involve the chain rule twice. Always double-check your sign and don't forget the factor that comes from differentiating the inner function.
16. Given $f(x) = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$, find $f'(\frac{1}{2})$.
    Answer: $\frac{\sqrt{3}}{3}$ This problem tests your ability to differentiate composite functions involving inverse trigonometric functions. When you see an arctangent function with a complex argument, you'll need to use the chain rule along with the derivative formula for arctangent. To find $f'(x)$, start with the derivative of arctangent: $\frac{d}{dx}[\arctan(u)] = \frac{1}{1+u^2} \cdot u'$. Here, $u = \frac{x}{1+\sqrt{1-x^2}}$, so you need to find $u'$ using the quotient rule. Let $g(x) = 1+\sqrt{1-x^2}$, so $u = \frac{x}{g(x)}$. First, $g'(x) = \frac{-x}{\sqrt{1-x^2}}$. Using the quotient rule: $u' = \frac{g(x) \cdot 1 - x \cdot g'(x)}{[g(x)]^2} = \frac{1+\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}}{(1+\sqrt{1-x^2})^2}$ After simplification, $u' = \frac{1}{\sqrt{1-x^2}(1+\sqrt{1-x^2})}$. At $x = \frac{1}{2}$: $u = \frac{1/2}{1+\sqrt{3}/2} = \frac{1}{2+\sqrt{3}} = \frac{\sqrt{3}}{3}$ (after rationalizing). Therefore: $f'(1/2) = \frac{1}{1+(\sqrt{3}/3)^2} \cdot \frac{1}{\sqrt{3/4}(1+\sqrt{3}/2)} = \frac{\sqrt{3}}{3}$ Answer C is correct. Answer A ($\frac{1}{2}$) likely comes from incorrectly evaluating the derivative. Answer B ($\frac{2\sqrt{3}}{3}$) suggests an error in simplifying the chain rule calculation. Answer D ($1$) might result from misapplying the arctangent derivative formula. Remember: when differentiating inverse trig functions with complex arguments, work methodically through the chain rule and double-check your algebra at each step.
17. Let $f(x) = \arcsin(x^3)$ and $g(x) = -\arccos(x^3)$ for $x \in (-1, 1)$. Which of the following statements is true?
    Answer: $f'(x) = g'(x)$ We compute the derivatives of both functions using the chain rule. For $f(x)$, let $u = x^3$. Then $f'(x) = \frac{1}{\sqrt{1-(x^3)^2}} \cdot 3x^2 = \frac{3x^2}{\sqrt{1-x^6}}$. For $g(x)$, we have $g'(x) = -(-\frac{1}{\sqrt{1-(x^3)^2}} \cdot 3x^2) = \frac{3x^2}{\sqrt{1-x^6}}$. Since the resulting expressions for $f'(x)$ and $g'(x)$ are identical for all $x$ in the domain, we have $f'(x) = g'(x)$. This reflects the general identity that $\frac{d}{dx}\arcsin(u) = -\frac{d}{dx}\arccos(u)$.
18. Let $f(x) = \arctan(x) + \arctan(\frac{1}{x})$ for $x \neq 0$. Find $f'(x)$.
    Answer: $0$ When you encounter a function involving inverse trigonometric expressions like this, consider whether there might be a special relationship that simplifies the derivative unexpectedly. To find $f'(x)$, you need to differentiate each term separately. For $\arctan(x)$, the derivative is $\frac{1}{1+x^2}$. For $\arctan(\frac{1}{x})$, you'll need the chain rule: the derivative of $\arctan(u)$ is $\frac{1}{1+u^2} \cdot u'$, where $u = \frac{1}{x}$ and $u' = -\frac{1}{x^2}$. This gives you $\frac{1}{1+(\frac{1}{x})^2} \cdot (-\frac{1}{x^2}) = \frac{1}{\frac{x^2+1}{x^2}} \cdot (-\frac{1}{x^2}) = \frac{x^2}{x^2+1} \cdot (-\frac{1}{x^2}) = -\frac{1}{x^2+1}$. Therefore, $f'(x) = \frac{1}{1+x^2} + (-\frac{1}{1+x^2}) = 0$. Choice A ($\frac{\pi}{2}$) represents a constant that might appear if you incorrectly tried to evaluate the original function rather than differentiate it. Choice B ($\frac{2}{1+x^2}$) would result from incorrectly assuming both terms have the same derivative. Choice C suggests you made sign errors or confused the derivatives of different inverse trig functions. The key insight here is recognizing when terms cancel completely. Always work through chain rule problems systematically, and don't assume complex-looking functions can't have simple derivatives. Sometimes the most elegant mathematical relationships produce surprisingly clean results.
19. Let $f(x) = \sin(x)$ and $g(x) = \arctan(x)$. If $h(x) = (g \circ f)(x)$, find $h'(\frac{\pi}{4})$.
    Answer: $\frac{\sqrt{2}}{3}$ First, write the composite function $h(x) = g(f(x)) = \arctan(\sin(x))$. To find its derivative, we use the chain rule: $h'(x) = \frac{1}{1+(\sin(x))^2} \cdot \cos(x)$. Now, we evaluate this derivative at $x = \frac{\pi}{4}$. We know that $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. Substituting these values: $h'(\frac{\pi}{4}) = \frac{1}{1+(\frac{\sqrt{2}}{2})^2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{1+\frac{2}{4}} \cdot \frac{\sqrt{2}}{2} = \frac{1}{1+\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} = \frac{1}{\frac{3}{2}} \cdot \frac{\sqrt{2}}{2} = \frac{2}{3} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{3}$.
20. Let $f(x) = \arctan(e^{2x})$. What is $f'(x)$?
    Answer: $\frac{2e^{2x}}{1+e^{4x}}$ This problem requires the chain rule along with the derivative of the inverse tangent function. Let $u = e^{2x}$. Then $f(u) = \arctan(u)$. The chain rule states that $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$. The derivative of $\arctan(u)$ with respect to $u$ is $\frac{1}{1+u^2}$. The derivative of $u = e^{2x}$ with respect to $x$ is $\frac{du}{dx} = e^{2x} \cdot 2 = 2e^{2x}$. Multiplying these together gives $f'(x) = \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x} = \frac{2e^{2x}}{1+e^{4x}}$. - Distractor B is incorrect because it omits the factor of 2 from the chain rule applied to $e^{2x}$. - Distractor C incorrectly uses the derivative formula for $\arcsin(u)$ instead of $\arctan(u)$. - Distractor D incorrectly omits the $e^{2x}$ term from the numerator, which arises from the chain rule.

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