Free math worksheets

- 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 \(x\in\mathbb{R}\)

- 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 \(x\in\mathbb{R}\)

- 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. a) Differentiating \(\arccos u=\frac{\pi}{2}-\arcsin u\) introduces a negative sign. The student omitted that sign. 2. b) By the chain rule, \(F^{\prime}(x)=\frac{2}{\sqrt{1-4x^2}}\) and \(G^{\prime}(x)=-\frac{2}{\sqrt{1-4x^2}}\), for \(-\frac{1}{2}<x<\frac{1}{2}\). 3. c) 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) \(F^{\prime}(x)=\frac{2}{\sqrt{1-4x^2}}\) and \(G^{\prime}(x)=-\frac{2}{\sqrt{1-4x^2}}\) c) \(F^{\prime}(x)+G^{\prime}(x)=0\), 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.

- Use the restricted sine function to determine the y-coordinate of \(P\). - An inverse graph reflects each point by reversing its coordinates. - Find the original tangent slope from the derivative of sine. - For corresponding inverse points with nonzero slope, compare rise and run after reflection across \(y=x\).

1. a) Since \(f\left(\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), \(P=\left(\frac{\pi}{3},\frac{\sqrt{3}}{2}\right)\). Reflection across \(y=x\) reverses coordinates, so \(Q=\left(\frac{\sqrt{3}}{2},\frac{\pi}{3}\right)\). 2. b) Since \(f^{\prime}(x)=\cos x\), the slope at \(P\) is \(f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1}{2}\). 3. c) At corresponding inverse points, the nonzero tangent slopes are reciprocals. Therefore, \(g^{\prime}\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{1/2}=2\). 4. d) Reflecting a nonvertical tangent line across \(y=x\) interchanges horizontal and vertical changes, so a slope of \(\frac{1}{2}\) becomes \(2\). The displayed tangent segments show this reciprocal steepness.

a) \(P=\left(\frac{\pi}{3},\frac{\sqrt{3}}{2}\right)\), \(Q=\left(\frac{\sqrt{3}}{2},\frac{\pi}{3}\right)\) b) \(\frac{1}{2}\) c) \(2\) d) Reflection across \(y=x\) swaps rise and run, so the nonzero slopes are reciprocals.