6.9 Calculus of the hyperbolic functions  (Page 3/5)

[T] Find expressions for $\text{cosh}x+\text{sinh}x$ and $\text{cosh}x-\text{sinh}x.$ Use a calculator to graph these functions and ensure your expression is correct.

From the definitions of $\text{cosh}\left(x\right)$ and $\text{sinh}\left(x\right),$ find their antiderivatives.

Show that $\text{cosh}\left(x\right)$ and $\text{sinh}\left(x\right)$ satisfy $y\text{″}=y.$

Use the quotient rule to verify that $\text{tanh}\left(x\right)\prime ={\text{sech}}^2\left(x\right).$

Derive ${\text{cosh}}^2\left(x\right)+{\text{sinh}}^2\left(x\right)=\text{cosh}\left(2x\right)$ from the definition.

Take the derivative of the previous expression to find an expression for $\text{sinh}\left(2x\right).$

Prove $\text{sinh}\left(x+y\right)=\text{sinh}\left(x\right)\text{cosh}\left(y\right)+\text{cosh}\left(x\right)\text{sinh}\left(y\right)$ by changing the expression to exponentials.

Take the derivative of the previous expression to find an expression for $\text{cosh}\left(x+y\right).$

[T] $\text{cosh}\left(3x+1\right)$

[T] $\text{sinh}\left(x^2\right)$

[T] $\frac{1}{\text{cosh}\left(x\right)}$

[T] $\text{sinh}\left(\text{ln}\left(x\right)\right)$

[T] ${\text{cosh}}^2\left(x\right)+{\text{sinh}}^2\left(x\right)$

[T] ${\text{cosh}}^2\left(x\right)-{\text{sinh}}^2\left(x\right)$

[T] $\text{tanh}\left(\sqrt{x^2+1}\right)$

[T] $\frac{1+\text{tanh}\left(x\right)}{1-\text{tanh}\left(x\right)}$

[T] ${\text{sinh}}^6\left(x\right)$

[T] $\text{ln}\left(\text{sech}\left(x\right)+\text{tanh}\left(x\right)\right)$

$\text{cosh}\left(2x+1\right)$

$\text{tanh}\left(3x+2\right)$

$x\text{cosh}\left(x^2\right)$

$3x^3\text{tanh}\left(x^4\right)$

${\text{cosh}}^2\left(x\right)\text{sinh}\left(x\right)$

${\text{tanh}}^2\left(x\right){\text{sech}}^2\left(x\right)$

$\frac{\text{sinh}\left(x\right)}{1+\text{cosh}\left(x\right)}$

$\text{coth}\left(x\right)$

$\text{cosh}\left(x\right)+\text{sinh}\left(x\right)$

${\left(\text{cosh}\left(x\right)+\text{sinh}\left(x\right)\right)}^n$

${\text{tanh}}^{\mathrm{-1}}\left(4x\right)$

${\text{sinh}}^{\mathrm{-1}}\left(x^2\right)$

${\text{sinh}}^{\mathrm{-1}}\left(\text{cosh}\left(x\right)\right)$

${\text{cosh}}^{\mathrm{-1}}\left(x^3\right)$

${\text{tanh}}^{\mathrm{-1}}\left(\text{cos}\left(x\right)\right)$

$e^{{\text{sinh}}^{\mathrm{-1}}\left(x\right)}$

$\text{ln}\left({\text{tanh}}^{\mathrm{-1}}\left(x\right)\right)$

${\displaystyle \int \frac{dx}{4-x^2}}$

${\displaystyle \int \frac{dx}{a^2-x^2}}$

${\displaystyle \int \frac{dx}{\sqrt{x^2+1}}}$

${\displaystyle \int \frac{xdx}{\sqrt{x^2+1}}}$

${\displaystyle \int -\frac{dx}{x\sqrt{1-x^2}}}$

${\displaystyle \int \frac{e^x}{\sqrt{e^{2x}-1}}}$

${\displaystyle \int -\frac{2x}{x^4-1}}$

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

Derive the previous expression for $v(t)$ by integrating $\frac{dv}{g-v^2}=dt.$

[T] Estimate how far a body has fallen in $12$ seconds by finding the area underneath the curve of $v(t).$

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

Integrate $dy\text{/}dx=\text{sinh}(cx)$ to find the cable height $y(x)$ if $y(0)=1\text{/}c.$

Sketch the cable and determine how far down it sags at $x=0.$