📝 Derivatives integrals inverse hyperbolic functions (31 MCQs)

📖 Explanation: To find the derivative, use the chain rule. The derivative of $\sinh^{-1}(u)$ with respect to x is $\frac{1}{\sqrt{1+u^2}} \frac{du}{dx}$. Here, $u = 3x^2$, so $du/dx = 6x$. Substituting gives $\frac{1}{\sqrt{1+(3x^2)^2}} \cdot 6x = \frac{6x}{\sqrt{1+9x^4}}$. Option B misses the factor from the derivative of the inside function, option C uses the derivative formula for $\cosh^{-1}$, and option D is for the derivative of $\cosh^{-1}$.

📖 Explanation: The derivative of $\cosh^{-1}(u)$ is $\frac{1}{\sqrt{u^2-1}}\frac{du}{dx}$. With $u=e^x$, $du/dx=e^x$. This gives $\frac{e^x}{\sqrt{e^{2x}-1}}$. The domain of $\cosh^{-1}(u)$ requires $u>1$, which is satisfied here for $x>0$. Option B is the derivative of $\sinh^{-1}(e^x)$, and option C misses the chain rule factor for the derivative of $e^x$.

📖 Explanation: The derivative of $\tanh^{-1}(u)$ is $\frac{1}{1-u^2}\frac{du}{dx}$. With $u = \cos x$, $du/dx = -\sin x$. Thus, the expression is $\frac{-\sin x}{1-\cos^2 x} = \frac{-\sin x}{\sin^2 x} = -\csc x$. Option C is an intermediate step. Option D is the same as C but with a sign change that negates the expression, making it incorrect. Understanding the identity $1-\cos^2 x = \sin^2 x$ is crucial for simplification.

📖 Explanation: The derivative of $\operatorname{sech}^{-1}(u)$ is $-\frac{1}{u\sqrt{1-u^2}}\frac{du}{dx}$. With $u = x^{1/2}$, $du/dx = \frac{1}{2\sqrt{x}}$. Substituting and simplifying: $-\frac{1}{x^{1/2}\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2x\sqrt{1-x}}$. Option A misses the negative sign, which is critical for $\operatorname{sech}^{-1}$. Option C is an unsimplified form that lacks the $x$ in the denominator, and option D uses the incorrect formula for $\operatorname{sech}^{-1}$.

📖 Explanation: The student's error lies in using the derivative formula for $\sin^{-1}$ instead of $\sinh^{-1}$. The derivative of $\sinh^{-1}(u)$ is $\frac{1}{\sqrt{1+u^2}}\frac{du}{dx}$, not $\frac{1}{\sqrt{1-u^2}}$. Here, $u=\tan x$, so the correct derivative is $\frac{\sec^2 x}{\sqrt{1+\tan^2 x}}$. This can be simplified to $|\cos x| \sec^2 x$ or $\sec x |\sec x|$, but the key is the plus sign in the denominator. Understanding the distinction between trigonometric and hyperbolic inverse derivatives is essential.

📖 Explanation: The integral formula is $\int \frac{du}{u\sqrt{a^2+u^2}} = -\frac{1}{a}\operatorname{sech}^{-1}\left|\frac{u}{a}\right| + C$. Here, $a=1$ and $u=2x$, so $du = 2dx$. This gives $\int \frac{dx}{x\sqrt{1+4x^2}} = -\operatorname{sech}^{-1}(|2x|) + C$. Option A is incorrect as it lacks the negative sign and absolute value. Option B is missing the absolute value. The absolute value is crucial because the domain of $\operatorname{sech}^{-1}$ is (0,1], and $2x$ could be negative.

📖 Explanation: The integral of $\frac{1}{\sqrt{x^2-a^2}}$ is $\cosh^{-1}(x/a) + C$ for $x>a$. Here, $a=3$, so the integral is $\cosh^{-1}(x/3) + C$. This is a direct Medium of a standard integral formula. Option A is the formula for $\int \frac{dx}{\sqrt{x^2+a^2}}$, and option C has an extra factor of $1/3$ which is incorrect. The integral of $1/\sqrt{x^2-a^2}$ is also $\ln(x+\sqrt{x^2-a^2}) + C$, but the hyperbolic form is $\cosh^{-1}$.

📖 Explanation: This requires the chain rule. The derivative of $\ln(u)$ is u'/u. Here, $u = \cosh^{-1}x$, and its derivative is $1/\sqrt{x^2-1}$. Therefore, the derivative is $\frac{1}{\cosh^{-1}x} \cdot \frac{1}{\sqrt{x^2-1}} = \frac{1}{\cosh^{-1}x \sqrt{x^2-1}}$. Option A is correct. Option B has an extra $1/x$. Option C would be the derivative of $\ln x$. The problem tests the ability to apply the chain rule systematically with inverse hyperbolic functions.

📖 Explanation: This is a standard integral $\int \frac{dt}{t^2+1} = \tan^{-1}(t) + C$. Evaluating from 0 to $\sqrt{3}$, we get $\tan^{-1}(\sqrt{3}) - \tan^{-1}(0) = \pi/3 - 0 = \pi/3$. This problem is a trap for those who might try to use hyperbolic functions. The denominator is $t^2+1$, not $1-t^2$ or $t^2-1$, so the inverse trigonometric tangent is the correct antiderivative. The hyperbolic tangent inverse would be for $\int \frac{dt}{1-t^2}$.

📖 Explanation: The derivative of $\coth^{-1}(u)$ is $\frac{1}{1-u^2}\frac{du}{dx}$, valid for $|u|>1$. Here, $u=\sec x$, $du/dx = \sec x \tan x$. So $\frac{dy}{dx} = \frac{\sec x \tan x}{1-\sec^2 x}$. Since $1-\sec^2 x = -\tan^2 x$, this simplifies to $-\frac{\sec x \tan x}{\tan^2 x} = -\frac{\sec x}{\tan x} = -\csc x$. The un-simplified form in option A is correct. Option B misses the negative sign from the chain rule? No, the derivative of $\sec x$ is positive. The negative comes from the denominator $1-\sec^2 x = -\tan^2 x$, which is negative for $|x| > \pi/2$. So the expression is negative. Option A captures this intermediate form. Option D is the simplified form, which is also correct, but option A is the direct result of the formula.

📖 Explanation: Complete the square: $x^2-4x+13 = (x-2)^2 + 9$. So the integral becomes $\int \frac{dx}{\sqrt{(x-2)^2 + 3^2}}$, which is of the form $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(u/a) + C$. Here, $u=x-2, a=3$. Thus, the integral is $\sinh^{-1}\left(\frac{x-2}{3}\right)+C$. Option B would be correct if the denominator was $\sqrt{u^2-a^2}$. Option C misses the division by 3, and option D has an extra factor of $1/3$.

📖 Explanation: The derivative of $\tanh^{-1}(u)$ is $\frac{1}{1-u^2}\frac{du}{dx}$. With $u = \sin x$, $du/dx = \cos x$. So the derivative is $\frac{\cos x}{1-\sin^2 x} = \frac{\cos x}{\cos^2 x} = \sec x$. The domain restriction ensures $\cos x > 0$ and $|\sin x| < 1$, so the expression is well-defined. Option D is an intermediate step but not simplified. Option B is $\cos x$, which is incorrect, and option C is the derivative of $\tan x$.

📖 Explanation: First, find the derivative: y&#039; = \frac{1}{\sqrt{1+(x/2)^2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{1+x^2/4}}. At $x=2$, y&#039;(2) = \frac{1}{2\sqrt{1+1}} = \frac{1}{2\sqrt{2}}. This is the slope of the tangent line. Option B is the derivative of $\sinh^{-1}(x)$. Option C is a common mistake of forgetting the square root. Option D is the value of the derivative if the denominator were $\sqrt{x^2+1}$ instead of $\sqrt{1+x^2/4}$.

📖 Explanation: Using the product rule: y&#039; = \operatorname{sech}^{-1}x + x \cdot \left(-\frac{1}{x\sqrt{1-x^2}}\right) = \operatorname{sech}^{-1}x - \frac{1}{\sqrt{1-x^2}}. At $x=0.5$, y&#039;(0.5) = \operatorname{sech}^{-1}(0.5) - \frac{1}{\sqrt{1-0.25}} = \operatorname{sech}^{-1}(0.5) - \frac{1}{\sqrt{0.75}}. Option A incorrectly leaves the $x$ term in the product rule. Option B is missing the $x$ in the product rule? No, option B has $\operatorname{sech}^{-1}(0.5) - 1/\sqrt{3}$, which is the correct simplified form. Option C is the unsimplified form which is also correct. Option D has an incorrect coefficient for the $\operatorname{sech}^{-1}$ term. I will choose C as the direct Medium.

📖 Explanation: The derivative of $\operatorname{csch}^{-1}(u)$ is $-\frac{1}{|u|\sqrt{1+u^2}}\frac{du}{dx}$. With $u = \tan x$, $du/dx = \sec^2 x$. Thus, $\frac{dy}{dx} = -\frac{\sec^2 x}{|\tan x|\sqrt{1+\tan^2 x}}$. Option A misses the negative sign, which is a common error. Option C has a $\sec x$ instead of $\sec^2 x$, and option D misses the absolute value, which is important for the domain of $\operatorname{csch}^{-1}$. The absolute value ensures the expression is valid for all $x$ where $\tan x \neq 0$.

📖 Explanation: The derivative of $\cosh^{-1}(u)$ is $\frac{1}{\sqrt{u^2-1}}\frac{du}{dx}$. Here, $u = \sec x$, $du/dx = \sec x \tan x$. So f&#039;(x) = \frac{\sec x \tan x}{\sqrt{\sec^2 x - 1}} = \frac{\sec x \tan x}{\sqrt{\tan^2 x}} = \frac{\sec x \tan x}{|\tan x|}. Since $x \in (0, \pi/2)$, $\tan x > 0$, so f&#039;(x) = \sec x. Option A is $\tan x$, option B is the derivative of $\sec x$, and option D is $\tan x \sec x$. The simplification step is crucial here. The absolute value disappears because the domain ensures $\tan x > 0$.

📖 Explanation: Complete the square: $x^2+2x+2 = (x+1)^2+1$. The integral is $\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C$. This is a standard inverse trigonometric integral. Option A is for $\int \frac{dx}{a^2-x^2}$, option B is for $\int \frac{dx}{\sqrt{1-x^2}}$, and option D is for $\int \frac{dx}{\sqrt{x^2-1}}$. The problem tests the ability to recognize the form $1/(u^2+a^2)$.

📖 Explanation: The derivative is $\frac{1}{\sqrt{1+\tan^2 x}} \cdot \sec^2 x$. This simplifies to $\frac{\sec^2 x}{|\sec x|} = |\sec x|$ for appropriate domains, but the unsimplified form is option A. Option B is the derivative of $\sin^{-1}(\tan x)$. Option C misses the chain rule factor, and option D is the derivative of $\tan^{-1}(\tan x)$.

📖 Explanation: Let $u = \sqrt{x^2+1}$, $du/dx = x/\sqrt{x^2+1}$. The derivative is $\frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx} = \frac{1}{\sqrt{x^2}} \cdot \frac{x}{\sqrt{x^2+1}} = \frac{x}{|x|\sqrt{x^2+1}}$. For $x>0$, this is $1/\sqrt{x^2+1}$. The unsimplified form is option A. Option B is the derivative of $\sinh^{-1}(x)$. Option C is the simplified form for $x>0$. Option D is for $\cosh^{-1}(x)$. The absolute value is often missed.

📖 Explanation: The integral is of the form $\int \frac{dx}{\sqrt{a^2x^2+b^2}}$. Let $u = 2x$, $du = 2dx$. Then the integral becomes $\frac{1}{2}\int \frac{du}{\sqrt{u^2+3^2}} = \frac{1}{2}\sinh^{-1}(u/3) + C = \frac{1}{2}\sinh^{-1}(2x/3) + C$. Option B has the a and b flipped. Option C is missing the $1/2$ factor. Option D uses $\cosh^{-1}$, which is for $\sqrt{u^2-a^2}$.

📖 Explanation: The derivative of $\tanh^{-1}(u)$ is $\frac{1}{1-u^2}\frac{du}{dx}$. With $u = \ln x$, $du/dx = 1/x$. Thus, y&#039; = \frac{1}{x(1-(\ln x)^2)}. Option B is the derivative of $\tan^{-1}(\ln x)$. Option C misses the $1/x$ factor. Option D has the wrong denominator form.

📖 Explanation: The derivative of $\coth^{-1}(u)$ is $\frac{1}{1-u^2}\frac{du}{dx}$. With $u = x^2$, $du/dx = 2x$, and $u^2 = x^4$. So y&#039; = \frac{2x}{1-x^4}. Since $x>1$, $1-x^4 < 0$, so the expression is negative. Option A is the direct result. Option B is the derivative of $\coth^{-1}(x)$? No, it's $\frac{1}{1-u^2}$ with a sign error. Option C is for $\coth^{-1}(x)$. Option D is the negative of the correct expression, which would be correct if the formula was $-1/(1-u^2)$.

📖 Explanation: The integrand $\frac{1}{1-x^2}$ has a vertical asymptote at $x=1$ and $x=-1$, so the integral is improper. The integral $\int_{-1}^{1} \frac{dx}{1-x^2}$ diverges to infinity. This is a common trap. The antiderivative is $\tanh^{-1}(x)$, but it is only valid for $|x|<1$. At the endpoints, the function is undefined. Option A is the antiderivative evaluated at the endpoints, but it's not a finite value. Option C is incorrect. Option D is also infinite. The correct answer is that the integral is divergent.

📖 Explanation: Let $u = \frac{1-x}{1+x}$. Then $du/dx = \frac{-(1+x)-(1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2}$. The derivative is $\frac{1}{\sqrt{1+u^2}} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1+(\frac{1-x}{1+x})^2}} \cdot \frac{-2}{(1+x)^2}$. Simplifying: $\sqrt{1+(\frac{1-x}{1+x})^2} = \sqrt{\frac{(1+x)^2+(1-x)^2}{(1+x)^2}} = \frac{\sqrt{(1-x)^2+(1+x)^2}}{|1+x|}$. So the derivative is $-\frac{2}{(1+x)^2} \cdot \frac{|1+x|}{\sqrt{(1-x)^2+(1+x)^2}} = -\frac{2}{(1+x)\sqrt{(1-x)^2+(1+x)^2}}$ for $1+x>0$. Option A is this simplified form. Option B is unsimplified with an extra factor of $1/(1+x)$. Option C has the wrong sign. Option D misses the square root term.