📝 Derivatives integrals of hyperbolic functions (26 MCQs)

📖 Explanation: The derivative of $\sinh(u)$ is $\cosh(u) \cdot \frac{du}{dx}$. Here, $u = x^2$, so $\frac{du}{dx} = 2x$. Thus, the derivative is $2x \cosh(x^2)$. Options B, C, and D incorrectly apply the chain rule or the derivative formula.

📖 Explanation: The integral of $\sinh(ax)$ is $\frac{1}{a} \cosh(ax) + C$. For $a = 3$, the answer is $\frac{1}{3} \cosh(3x) + C$. Option B is the derivative of $\sinh(3x)$, and Option D would be the integral of $-\sinh(3x)$. Option C is missing the division by 3.

📖 Explanation: The integral of $\operatorname{sech}^2(u)$ is $\tanh(u) + C$. However, due to the chain rule in reverse, we must account for the derivative of $u = 2x$. Since $\frac{d}{dx}[\tanh(2x)] = 2 \operatorname{sech}^2(2x)$, the integral is $\frac{1}{2} \tanh(2x) + C$. Options B and C ignore the need to divide by the derivative of the inner function, and Option D is not a valid integration result.

📖 Explanation: Using the chain rule, $\frac{dy}{dx} = \frac{1}{\tanh x} \cdot \frac{d}{dx}[\tanh x] = \frac{\operatorname{sech}^2 x}{\tanh x}$. This can also be simplified to $\frac{1}{\sinh x \cosh x}$. Option B is the derivative of $\ln(\tanh x)$ without applying the chain rule, and Option C is the derivative of $\tanh x$. Option D is the reciprocal of the correct expression.

📖 Explanation: The derivative of the inverse hyperbolic cosine is a standard formula: $\frac{d}{dx}[\cosh^{-1} x] = \frac{1}{\sqrt{x^2 - 1}}$. Option B is the derivative of $\sin^{-1} x$. Option D is the derivative of $\sinh^{-1} x$. Option C is missing the square root. This formula is valid for $x > 1$, as $\cosh^{-1} x$ is only defined for $x \ge 1$.

📖 Explanation: The integral is of the form $\int \frac{du}{\sqrt{u^2 + a^2}}$. Here, $u = 3x$, so $du = 3 dx$. Thus, $\int \frac{dx}{\sqrt{(3x)^2 + 4^2}} = \frac{1}{3} \sinh^{-1}\left(\frac{3x}{4}\right) + C$. Option B misses the $\frac{1}{3}$ factor. Option C incorrectly identifies the coefficients. While Option D is a logarithmic form, it lacks the correct multiplicative constant and is equivalent to $\sinh^{-1}\left(\frac{3x}{4}\right) + \ln(4)$, not $\frac{1}{3} \sinh^{-1}\left(\frac{3x}{4}\right)$.

📖 Explanation: The formula $\int \frac{du}{a^2 - u^2} = \frac{1}{a} \tanh^{-1}(u/a)$ is only valid when $|u| < a$. Since the student did not specify the domain of $x$, the answer is incomplete. For $|x| > 2$, the correct antiderivative is $\frac{1}{2} \coth^{-1}(x/2) + C$. Option B has the derivative wrong. Option C has the incorrect constant. Option D is only true for $|x| > 2$, so it is not universally correct.

📖 Explanation: The derivative of $\operatorname{sech}^{-1}(u)$ is $-\frac{1}{u \sqrt{1 - u^2}} \cdot \frac{du}{dx}$. Here, $u = e^x$ and $\frac{du}{dx} = e^x$. This gives $-\frac{1}{e^x \sqrt{1 - e^{2x}}} \cdot e^x = -\frac{1}{\sqrt{1 - e^{2x}}}$. Option A is the final simplification. Option B misses the $1/u$ term. Option D has the correct format but is missing the negative sign.

📖 Explanation: This requires the chain rule and the power rule. First, the derivative of $u^3$ is $3u^2 \cdot \frac{du}{dx}$. Here, $u = \sinh(2x)$, so $\frac{du}{dx} = 2 \cosh(2x)$. Multiplying gives $3 \sinh^2(2x) \cdot 2 \cosh(2x) = 6 \sinh^2(2x) \cosh(2x)$. Option B is missing the factor of 2 from the derivative of $2x$. Option C incorrectly applies the derivative of $e^x$.

📖 Explanation: The arc length formula is $L = \int_a^b \sqrt{1 + (dy/dx)^2} \, dx$. For $y = 3 \cosh(x/3)$, $dy/dx = \sinh(x/3)$. Thus, the integrand is $\sqrt{1 + \sinh^2(x/3)} = \cosh(x/3)$. Option A is the simplified correct integral. Option B is the same integrand before simplification. Option C has the wrong derivative (missing the factor from the chain rule). Option D has an extra factor of 3.

📖 Explanation: The derivative of $\tanh^{-1}(u)$ is $\frac{1}{1 - u^2} \cdot \frac{du}{dx}$. Here, $u = \sqrt{x}$, so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$. The final expression is $\frac{1}{1 - x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1 - x)}$. Note that Option A and B are identical (a typo in the options list). Option C misses the chain rule term. Option D has a $+x$ instead of $-x$.

📖 Explanation: The integral of $\tanh x$ is $\ln(\cosh x) + C$. Evaluating from 0 to $\ln 2$: $\ln(\cosh(\ln 2)) - \ln(\cosh 0)$. $\cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + 1/2}{2} = \frac{5}{4}$. $\cosh 0 = 1$. Thus, the result is $\ln(5/4)$. Option B is the negative of the correct answer. Option C ignores the integral formula. Option D is a numeric approximation that is not exact.

📖 Explanation: This is a classic Medium of the chain rule. $\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \sinh x = \tanh x$. Option B is the inner function. Option C is not a derivative of $\ln(\cosh x)$. Option D would be the derivative of $\sinh^2 x$ or $\cosh^2 x$.

📖 Explanation: First derivative: y&#039; = \sinh x + x \cosh x. Second derivative: y&#039;&#039; = \cosh x + (\cosh x + x \sinh x) = 2 \cosh x + x \sinh x. Option B is the first derivative. Option C is a misMedium of the product rule. Option D is an incorrect coefficient for the hyperbolic term.

📖 Explanation: The derivative of $\operatorname{csch}^{-1} x$ is a standard formula: $-\frac{1}{|x| \sqrt{1 + x^2}}$. The absolute value is crucial because the domain of $\operatorname{csch}^{-1} x$ includes negative values, and the derivative must be negative for all $x \neq 0$. Option A is missing the absolute value. Option C is the derivative of $\sec^{-1} x$. Option D is the correct formula. Option B is the derivative of $\operatorname{sech}^{-1} x$.

📖 Explanation: The area under the curve is $\int_0^{\ln 2} \sinh x \, dx = [\cosh x]_0^{\ln 2} = \cosh(\ln 2) - \cosh 0 = \cosh(\ln 2) - 1$. Option B is the integral of $\cosh x$. Option C is the integrand for a different area. Option D is irrelevant. Calculating $\cosh(\ln 2) = (2 + 0.5)/2 = 1.25$, so the area is 0.25.

📖 Explanation: The integral $\int \frac{dx}{\sqrt{x^2 - a^2}}$ is a standard form that results in $\cosh^{-1}(x/a)$. The derivative of $\operatorname{sech}^{-1}(x/2)$ is $-\frac{1}{x \sqrt{1 - x^2/4}}$, which is not the integrand. The student confused this with the integral $\int \frac{dx}{x \sqrt{a^2 - x^2}}$, which yields $-\frac{1}{a} \operatorname{sech}^{-1}(x/a)$. This is a common misconception. The correct answer for $\int \frac{dx}{\sqrt{x^2 - 4}}$ is $\cosh^{-1}(x/2)$.

📖 Explanation: Let $u = \operatorname{sech} x$. Then $y = u^2$, so $\frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2 \operatorname{sech} x \cdot (-\operatorname{sech} x \tanh x) = -2 \operatorname{sech}^2 x \tanh x$. Option B is missing one $\operatorname{sech} x$. Option C has the wrong sign. Option D has the wrong power on $\tanh x$.

📖 Explanation: Let $u = e^x$, so $du = e^x dx$. The integral becomes $\int \sinh(u) \, du = \cosh(u) + C = \cosh(e^x) + C$. Option B is the integral of $\cosh(e^x)$. Option C is the same as B. Option D has the wrong sign for the integral of $\sinh(u)$.

📖 Explanation: Since $\sinh x$ is strictly increasing, its inverse $\sinh^{-1} x$ is also strictly increasing. The range of $\sinh x$ is $(-\infty, \infty)$, so the domain of its inverse is all real numbers. It does not have vertical or horizontal asymptotes; it continues to increase linearly as $x \to \pm\infty$. Option A is incorrect because there are no vertical asymptotes. Option B is incorrect because there are no horizontal asymptotes. Option D is wrong because the function is increasing, not decreasing.

📖 Explanation: Let $u = \coth x$. Then $du = -\operatorname{csch}^2 x \, dx$. The integral becomes $\int u \cdot (-\operatorname{csch}^2 x) \, dx = -\int u \, du = -\frac{u^2}{2} + C = -\frac{1}{2} \coth^2 x + C$. Option B is the integral of $\operatorname{csch}^2 x \coth x$ if one mistakenly uses $u = \operatorname{csch} x$. Option C has the wrong sign. Option D is missing the constant of integration.