- Recall the derivative rule for \(\arcsin x\). - Substitute the requested input only after writing the derivative formula. - Check that the requested input lies inside the derivative's domain.

1. The inverse-sine derivative is \(f^{\prime}(x)=\frac{1}{\sqrt{1-x^2}}\) for \(-1<x<1\). 2. Substituting \(x=0\) gives \(f^{\prime}(0)=1\).

\(1\)

- Distinguish the inverse-tangent rule from the inverse-sine rule. - Examine whether the denominator of each candidate can become zero for real \(x\). - The domain of the derivative should agree with the behavior of the correct denominator.

1. The derivative rule for inverse tangent is \(\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}\). 2. Because \(1+x^2>0\) for every real \(x\), this derivative exists for all real \(x\).

\(\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}\), for \(x\in\mathbb{R}\)

- Do not confuse the derivative of sine with that of inverse sine. - Is inverse sine increasing or decreasing in the middle of its domain?

1. The derivative rule for inverse sine is \(\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}\) for \(-1<x<1\). 2. This matches option b).

b) \(\frac{1}{\sqrt{1-x^2}}\)

- Write the derivative rule for inverse cosine before substituting the input. - How does the sign of the derivative determine whether the curve rises or falls?

1. The derivative of inverse cosine is \(\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\). 2. At \(x=0\), the slope is \(-1\), so the graph is falling.

The slope is \(-1\); the graph is falling.

- Being able to evaluate a function does not always mean it has a derivative there. - What happens to the derivative formula as you approach the right endpoint of the domain?

1. The derivative formula is \(f'(x)=\frac{1}{\sqrt{1-x^2}}\) for \(-1<x<1\). 2. At \(x=1\), its denominator becomes \(0\), so there is no finite derivative there, even though \(f(1)\) exists.

No. The denominator of the derivative formula is zero at \(x=1\).

- Compare the graph's height at nearby x-values. - How would you describe a tangent line near \(x=0\)?

1. The graph moves downward as x increases, so its slopes are negative in the interior. 2. Therefore the derivative must be negative, consistent with \(-\frac{1}{\sqrt{1-x^2}}\).

Negative, because the curve decreases as x increases.

- Identify the expression serving as the input to \(\arcsin\). - Apply the inverse-sine derivative rule to that input. - Include the derivative of the inner linear expression.

1. Use the inverse-sine derivative with inner function \(u=3x\): \(\frac{d}{dx}(\arcsin u)=\frac{u^{\prime}}{\sqrt{1-u^2}}\). 2. Since \(u^{\prime}=3\), \(f^{\prime}(x)=\frac{3}{\sqrt{1-9x^2}}\). 3. This derivative is finite for \(-\frac{1}{3}<x<\frac{1}{3}\).

\(f^{\prime}(x)=\frac{3}{\sqrt{1-9x^2}}\), for \(-\frac{1}{3}<x<\frac{1}{3}\)

- Treat the entire quadratic expression as the input to \(\arctan\). - The inverse-tangent derivative has a denominator of the form \(1+u^2\). - Multiply by the derivative of the inner expression.

1. Let \(u=x^2+1\). Then \(u^{\prime}=2x\). 2. Use \(\frac{d}{dx}(\arctan u)=\frac{u^{\prime}}{1+u^2}\): \(g^{\prime}(x)=\frac{2x}{1+(x^2+1)^2}\). 3. The denominator is positive for every real \(x\), so the derivative exists for all real \(x\).

\(g^{\prime}(x)=\frac{2x}{1+(x^2+1)^2}\), for \(x\in\mathbb{R}\)

- Find the derivative of the inverse-tangent function before comparing slopes. - After substituting each input, compare the two fractions without rounding.

1. Differentiate: \(f'(x)=\frac{1}{1+x^2}\). 2. Evaluate: \(f'(0)=1\) and \(f'(1)=\frac{1}{2}\). 3. The tangent at \(x=0\) has the greater slope.

At \(x=0\): slope \(1\). At \(x=1\): slope \(\frac{1}{2}\). The slope is greater at \(x=0\).

- Identify the expression fed into inverse sine. - Multiply the inverse-sine derivative by the derivative of that expression. - The derivative's square-root denominator must be positive.

1. The inner expression is \(u(x)=\frac{x+1}{3}\), with derivative \(u'(x)=\frac{1}{3}\). 2. The chain rule gives \(f'(x)=\frac{1/3}{\sqrt{1-(\frac{x+1}{3})^2}}=\frac{1}{\sqrt{9-(x+1)^2}}\). 3. This derivative exists when \(-1<\frac{x+1}{3}<1\), or \(-4<x<2\).

\(f'(x)=\frac{1}{\sqrt{9-(x+1)^2}}\) for \(-4<x<2\).

- Keep the square inside the inverse-cosine derivative before simplifying. - Which sign belongs to the derivative of inverse cosine? - Where is the square root in the denominator nonzero?

1. The outer inverse-cosine derivative contributes a minus sign: \(-\frac{1}{\sqrt{1-(x^2)^2}}\). 2. Multiply by the inner derivative \(2x\), giving \(g'(x)=-\frac{2x}{\sqrt{1-x^4}}\). 3. The function requires \(0\leq x^2\leq1\), and the derivative denominator is positive for \(-1<x<1\).

\(g'(x)=-\frac{2x}{\sqrt{1-x^4}}\) for \(-1<x<1\).

- Keep the entire quadratic expression as the input to inverse tangent. - The denominator of an inverse-tangent derivative contains the square of that input. - Check whether the denominator can be zero.

1. The inner expression is \(u(x)=3x^2-1\), so \(u'(x)=6x\). 2. Use the inverse-tangent rule and chain rule: \(h'(x)=\frac{6x}{1+(3x^2-1)^2}\). 3. The denominator is always positive, so the derivative exists for all real \(x\).

\(h'(x)=\frac{6x}{1+(3x^2-1)^2}\), for all real \(x\).

- Find both the function value and slope at the given input. - Does the inverse-cosine graph rise or fall as its input increases? - Use the point and slope to form the tangent equation.

1. The point is \(\left(0,\frac{\pi}{2}\right)\), since \(f(0)=\arccos(0)=\frac{\pi}{2}\). 2. The chain rule gives \(f'(x)=-\frac{1}{\sqrt{4-x^2}}\), so \(f'(0)=-\frac{1}{2}\). 3. The tangent line is \(y-\frac{\pi}{2}=-\frac{1}{2}x\).

\(y=\frac{\pi}{2}-\frac{x}{2}\)

- In the inverse-tangent derivative rule, which quantity is squared? - Check the denominator of the student's result against the original function's input.

1. The derivative of inverse tangent has denominator \(1+u^2\) when the input is \(u\). 2. Here the input is \(2x\), and its square is \((2x)^2=4x^2\), not \(2x^2\). 3. Multiply by the inner derivative \(2\) to get \(q'(x)=\frac{2}{1+4x^2}\).

The input was squared incorrectly. \(q'(x)=\frac{2}{1+4x^2}\).

- Start from the inverse relationship \(\sin y=x\), not from a memorized derivative formula. - Use a Pythagorean identity to replace \(\cos y\) by an expression involving \(x\). - The principal range of \(\arcsin\) determines the sign of \(\cos y\). - Check where the final denominator is zero.

1. a) Differentiate \(\sin y=x\): \(\cos y\frac{dy}{dx}=1\). Therefore, \(\frac{dy}{dx}=\frac{1}{\cos y}\). 2. b) Since \(\sin y=x\), \(\cos^2 y=1-\sin^2 y=1-x^2\). On the principal range of \(\arcsin\), \(\cos y\ge0\), and for \(-1<x<1\), \(\cos y>0\). Hence \(\cos y=\sqrt{1-x^2}\). Therefore, \(\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}\) for \(-1<x<1\). 3. c) At \(x=\pm1\), the denominator \(\sqrt{1-x^2}\) is \(0\), so the derivative is not finite there.

a) \(\frac{dy}{dx}=\frac{1}{\cos y}\) b) \(\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}\), for \(-1<x<1\) c) At \(x=\pm1\), the denominator is \(0\), so there is no finite derivative value.

- Differentiate the identity relating inverse cosine and inverse sine. - Compare the two derivative rules before applying the common inner derivative. - A constant sum should have derivative \(0\).

1. Differentiating \(\arccos u=\frac{\pi}{2}-\arcsin u\) introduces a negative sign. Amina omitted that sign. 2. By the chain rule, \(F'(x)=\frac{2}{\sqrt{1-4x^2}}\) and \(G'(x)=-\frac{2}{\sqrt{1-4x^2}}\), for \(-\frac12<x<\frac12\). 3. Their derivatives add to \(0\), which is the derivative of the constant \(\frac{\pi}{2}\). This agrees with \(F+G=\frac{\pi}{2}\).

a) The derivative of \(\arccos u\) has a negative sign. b) \(F'(x)=\frac{2}{\sqrt{1-4x^2}}\) and \(G'(x)=-\frac{2}{\sqrt{1-4x^2}}\) c) \(F'(x)+G'(x)=0\), consistent with the derivative of \(\frac{\pi}{2}\).

- Identify the outermost algebraic structure before differentiating the inverse-trigonometric factor. - Keep the undifferentiated factor in each product-rule term. - Treat \(3x\) as the inner function of \(\arctan\).

1. a) The expression is a product, so use the product rule. Differentiating \(\arctan(3x)\) also requires the inverse-tangent derivative rule and the chain rule. 2. b) Differentiate each factor: \(p^{\prime}(x)=2x\arctan(3x)+x^2\left(\frac{3}{1+9x^2}\right)\). Therefore, \(p^{\prime}(x)=2x\arctan(3x)+\frac{3x^2}{1+9x^2}\).

a) Product rule, inverse-tangent derivative rule, and chain rule b) \(p^{\prime}(x)=2x\arctan(3x)+\frac{3x^2}{1+9x^2}\)

- Evaluate the inverse-sine function at the given x-value using an exact special angle. - Differentiate the inner fraction as part of the chain rule. - Use the exact derivative value as the slope in point-slope form.

1. At \(x=1\), \(f(1)=\arcsin\left(\frac{1}{2}\right)=\frac{\pi}{6}\), so the point is \(\left(1,\frac{\pi}{6}\right)\). 2. Differentiate using the inverse-sine rule and chain rule: \(f^{\prime}(x)=\frac{\frac{1}{2}}{\sqrt{1-\frac{x^2}{4}}}=\frac{1}{\sqrt{4-x^2}}\). 3. Thus \(f^{\prime}(1)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\). 4. The tangent line is \(y-\frac{\pi}{6}=\frac{\sqrt{3}}{3}(x-1)\).

Point: \(\left(1,\frac{\pi}{6}\right)\) Slope: \(\frac{\sqrt{3}}{3}\) Tangent line: \(y-\frac{\pi}{6}=\frac{\sqrt{3}}{3}(x-1)\)

- Start with the input restriction for \(\arcsin\), including its endpoint values. - Solve the resulting compound inequality in terms of \(x^2\) before converting it to x-intervals. - For the derivative domain, distinguish between allowing the inverse-sine input to equal \(\pm1\) and requiring the derivative denominator to stay nonzero.

1. a) The input to \(\arcsin\) must satisfy \(-1\le x^2-2\le1\). Adding \(2\) gives \(1\le x^2\le3\). Therefore, the domain of \(h\) is \([-\sqrt{3},-1]\cup[1,\sqrt{3}]\). 2. b) By the inverse-sine rule and chain rule, \(h^{\prime}(x)=\frac{2x}{\sqrt{1-(x^2-2)^2}}\). A finite derivative requires \(-1<x^2-2<1\), so \(1<x^2<3\). Thus, the derivative domain is \((-\sqrt{3},-1)\cup(1,\sqrt{3})\). 3. c) At \(x=\pm1\) or \(x=\pm\sqrt{3}\), the input to \(\arcsin\) is an endpoint value \(-1\) or \(1\), making the square-root denominator in the derivative formula equal to \(0\). The function is defined there, but its derivative is not finite there.

a) \([-\sqrt{3},-1]\cup[1,\sqrt{3}]\) b) \(h^{\prime}(x)=\frac{2x}{\sqrt{1-(x^2-2)^2}}\), with domain \((-\sqrt{3},-1)\cup(1,\sqrt{3})\) c) At all four endpoints, the inverse-sine derivative denominator is \(0\), so the function is defined but does not have a finite derivative there.

- Treat y as a function of x when differentiating cosine. - Use a trigonometric identity to replace sine by an expression involving cosine. - Which sign of sine is correct for angles between 0 and pi?

1. Differentiate \(\cos y=x\) with respect to \(x\): \(-\sin y\frac{dy}{dx}=1\). 2. Thus \(\frac{dy}{dx}=-\frac{1}{\sin y}\). 3. Since \(0<y<\pi\), \(\sin y>0\), and \(\sin y=\sqrt{1-\cos^2 y}=\sqrt{1-x^2}\). 4. Hence \(\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\).

\(\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}}\), for \(-1<x<1\).