쌍곡선함수의 미적분

Show that $\cosh (x)$ and $\sinh (x)$ satisfy ${y}^{''}=y.$

Prove $\sinh (x+y)=\sinh (x)\cosh (y)+\cosh (x)\sinh (y)$ by changing the expression to exponentials.

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation $dv/dt=g-{v}^{2}.$

Show that $v(t)=\sqrt{g}\,\tanh \left(\left(\sqrt{g}\right)t\right)$ satisfies this equation.

For the following exercises, use this scenario: A cable hanging under its own weight has a slope $S=dy/dx$ that satisfies $dS/dx=c\sqrt{1+{S}^{2}}.$ The constant $c$ is the ratio of cable density to tension.

Show that $S=\sinh (cx)$ satisfies this equation.

Prove the formula for the derivative of $y={\sinh }^{-1}(x)$ by differentiating $x=\sinh (y).$ (Hint: Use hyperbolic trigonometric identities.)

Prove the formula for the derivative of $y={\cosh }^{-1}(x)$ by differentiating $x=\cosh (y).$

(Hint: Use hyperbolic trigonometric identities.)

For the following exercises, solve each problem.

Prove that ${(\cosh (x)+\sinh (x))}^{n}=\cosh (nx)+\sinh (nx).$

Prove the expression for ${\sinh }^{-1}(x).$ Multiply $x=\sinh (y)=(1/2)({e}^{y}-{e}^{-y})$ by $2{e}^{y}$ and solve for $y.$ Does your expression match the textbook?

Prove the expression for ${\cosh }^{-1}(x).$ Multiply $x=\cosh (y)=(1/2)({e}^{y}+{e}^{-y})$ by $2{e}^{y}$ and solve for $y.$ Does your expression match the textbook?