Day 32

 

moment

 to compare the derivatives of the 

hyperbolic functions

 with the derivatives of th standard 

trigonometric functions

. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match:

ddxsinx=cosx$${\displaystyle \frac{d}{dx}}\sin x=\cos x$$

and

ddxsinhx=coshx.$${\displaystyle \frac{d}{dx}}\sinh x=\cosh x.$$

The derivatives of the cosine functions, however, differ in sign:

ddxcosx=−sinx,$${\displaystyle \frac{d}{dx}}\cos x=-\sin x,$$

but

ddxcoshx=sinhx.$${\displaystyle \frac{d}{dx}}\cosh x=\sinh x.$$

As we continue our examination of the 

hyperbolic functions

, we must be mindful of their similarities and differences to the standard 

trigonometric functions

. These 

differentiation

 formulas for the 

hyperbolic functions

 lead directly to the following integral formulas.

∫sinhudu$$\begin{array}{}\text{(6.9.1)}& \int \sinh udu& =\cosh u+C\text{(6.9.2)}& \int {\text{csch}}^{2}udu& =-\coth u+C\text{(6.9.3)}& \int \cosh udu& =\sinh u+C\text{(6.9.4)}& \int \text{sech}u\tanh udu& =-\text{sech }u+C-\text{csch}u+C\text{(6.9.5)}& \int {\text{sech }}^{2}udu& =\tanh u+C\text{(6.9.6)}& \int \text{csch}u\coth udu& =-\text{csch}u+C\end{array}$$

Example 6.9.1$\mathrm{6.9.}1$: Differentiating 

Hyperbolic Functions

Evaluate the following derivatives:

1. ddx(sinh(x2))${\displaystyle \frac{d}{dx}}(\sinh (x^2))$
2. ddx(coshx)2${\displaystyle \frac{d}{dx}}(\cosh x{)}^2$

Solution:

Using the formulas in Table 6.9.1$\mathrm{6.9.}1$ and the 

chain rule

, we get

1. ddx(sinh(x2))=cosh(x2)⋅2x${\displaystyle \frac{d}{dx}}(\sinh (x^2))=\cosh (x^2)\cdot 2x$
2. ddx(coshx)2=2coshxsinhx${\displaystyle \frac{d}{dx}}(\cosh x{)}^2=2\cosh x\sinh x$

Exercise 6.9.1$\mathrm{6.9.}1$

Evaluate the following derivatives:

1. ddx(tanh(x2+3x))${\displaystyle \frac{d}{dx}}(\tanh (x^2+3x))$
2. ddx(1(sinhx)2)${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{1}{(\sinh x{)}^2}}\right)$

Hint

Answer a

Answer b

Example 6.9.2$\mathrm{6.9.}2$: Integrals Involving 

Hyperbolic Functions

Evaluate the following integrals:

1. ∫xcosh(x2)dx${\displaystyle \int x\cosh (x^2)dx}$
2. ∫tanhxdx${\displaystyle \int \tanh xdx}$

Solution

We can use u$u$-substitution in both cases.

a. Let u=x2$u=x^2$. Then, du=2xdx$du=2xdx$ and

∫xcosh(x2)dx$$\begin{array}{rl}\int x\cosh (x^2)dx& =\int {\displaystyle \frac{1}{2}}\cosh udu\\ & ={\displaystyle \frac{1}{2}}\sinh u+C\\ & ={\displaystyle \frac{1}{2}}\sinh (x^2)+C.\end{array}$$

b. Let u=coshx$u=\cosh x$. Then, du=sinhxdx$du=\sinh xdx$ and

∫tanhxdx$$\begin{array}{rl}\int \tanh xdx& =\int {\displaystyle \frac{\sinh x}{\cosh x}}dx\\ & =\int {\displaystyle \frac{1}{u}}du\\ & =\ln |u|+C\\ & =\ln |\cosh x|+C.\end{array}$$

Note that coshx>0$\cosh x>0$ for all x$x$, so we can eliminate the absolute value signs and obtain

∫tanhxdx=ln(coshx)+C.$$\int \tanh xdx=\ln (\cosh x)+C.$$

Exercise 6.9.2$\mathrm{6.9.}2$

Evaluate the following integrals:

1. ∫sinh3xcoshxdx${\displaystyle \int {\sinh }^{3}x\cosh xdx}$
2. ∫sech 2(3x)dx${\displaystyle \int {\text{sech }}^2(3x)dx}$

Hint

Answer a

Answer b