Core Pure · Hyperbolic Functions · Year 2

Differentiating & Integrating Hyperbolic Functions

Standard derivatives and integrals of sinh, cosh, tanh and their inverses, plus hyperbolic substitution.

15 min read 4 worked examples

Differentiating hyperbolic functions

Theorem 6.4 — standard derivatives

$$\frac{d}{dx}(\sinh x)=\cosh x$$

$$\frac{d}{dx}(\cosh x)=\sinh x$$

$$\frac{d}{dx}(\tanh x)=\operatorname{sech}^2x$$

Notice there is no minus sign in the derivative of $\cosh x$, unlike $\dfrac{d}{dx}(\cos x)=-\sin x$ — this is one place Osborn's rule doesn't apply blindly.

ProofDerivative of cosh x

$\cosh x=\dfrac{e^x+e^{-x}}2$.

Differentiate term by term: $\dfrac{d}{dx}\left(\dfrac{e^x+e^{-x}}2\right)=\dfrac{e^x+(-1)e^{-x}}2=\dfrac{e^x-e^{-x}}2$.

The $e^x$ term differentiates to $e^x$; the $e^{-x}$ term differentiates to $-e^{-x}$, i.e. picks up a factor of $-1$.

This is exactly $\sinh x$ by definition.

∎

ProofDerivative of tanh x, via the quotient rule

$\tanh x=\dfrac{\sinh x}{\cosh x}$. By the quotient rule, $\dfrac{d}{dx}(\tanh x)=\dfrac{\cosh x\cdot\cosh x-\sinh x\cdot\sinh x}{\cosh^2x}$.

$=\dfrac{\cosh^2x-\sinh^2x}{\cosh^2x}$.

By the fundamental identity, the numerator is $1$, so this is $\dfrac1{\cosh^2x}=\operatorname{sech}^2x$.

∎

Example 1 · Chain rule with hyperbolic functions

Given $y=A\cosh5x+B\sinh5x$, where $A,B$ are constants, show that $\dfrac{d^2y}{dx^2}=25y$.

$\dfrac{dy}{dx}=5A\sinh5x+5B\cosh5x$.

$\dfrac{d^2y}{dx^2}=25A\cosh5x+25B\sinh5x=25(A\cosh5x+B\sinh5x)=25y$.

Derivatives of the inverse hyperbolic functions

$$\frac{d}{dx}(\operatorname{arsinh}x)=\frac1{\sqrt{x^2+1}}$$

$$\frac{d}{dx}(\operatorname{arcosh}x)=\frac1{\sqrt{x^2-1}}$$

$$\frac{d}{dx}(\operatorname{artanh}x)=\frac1{1-x^2}$$

ProofDerivative of arsinh x

Let $y=\operatorname{arsinh}x$, so $x=\sinh y$. Differentiate implicitly with respect to $x$: $1=\cosh y\cdot\dfrac{dy}{dx}$.

So $\dfrac{dy}{dx}=\dfrac1{\cosh y}$. Using $\cosh^2y-\sinh^2y=1$ and $\cosh y\geq0$ always: $\cosh y=\sqrt{1+\sinh^2y}=\sqrt{1+x^2}$.

So $\dfrac{dy}{dx}=\dfrac1{\sqrt{x^2+1}}$.

∎

Example 2 · Differentiating a composite involving arcosh

Find the equation of the tangent to $y=\operatorname{arcosh}(3x)$ at the point where $x=1$.

By the chain rule, $\dfrac{dy}{dx}=\dfrac3{\sqrt{(3x)^2-1}}$.

At $x=1$: $\dfrac{dy}{dx}=\dfrac3{\sqrt8}=\dfrac{3}{2\sqrt2}$, and $y=\operatorname{arcosh}3=\ln\left(3+\sqrt8\right)=\ln\left(3+2\sqrt2\right)$.

Tangent: $y-\ln\left(3+2\sqrt2\right)=\dfrac3{2\sqrt2}(x-1)$.

Integrating hyperbolic functions

Theorem 6.5 — standard integrals

$$\int\sinh x\,dx=\cosh x+c$$

$$\int\cosh x\,dx=\sinh x+c$$

$$\int\frac1{\sqrt{x^2+1}}\,dx=\operatorname{arsinh}x+c$$

$$\int\frac1{\sqrt{x^2-1}}\,dx=\operatorname{arcosh}x+c\;(x>1)$$

$$\int\frac1{1-x^2}\,dx=\operatorname{artanh}x+c\;(|x|<1)$$

Example 3 · Powers of hyperbolic functions

Find $\displaystyle\int\sinh^24x\,dx$.

Use $\cosh8x\equiv1+2\sinh^24x$, so $\sinh^24x=\dfrac{\cosh8x-1}2$.

$$\int\sinh^24x\,dx=\int\frac{\cosh8x-1}2\,dx=\frac1{16}\sinh8x-\frac x2+c$$

Method — hyperbolic substitution for √(x² − a²) and √(x² + a²)

Where a trig substitution would use $x=a\sin u$ (for $\sqrt{a^2-x^2}$), hyperbolic substitutions handle the other two square-root forms: use $x=a\sinh u$ for integrands involving $\sqrt{x^2+a^2}$, and $x=a\cosh u$ for integrands involving $\sqrt{x^2-a^2}$ — exploiting $\cosh^2u-\sinh^2u=1$ to simplify the square root.

Example 4 · Hyperbolic substitution

Using the substitution $x=3\sinh u$, find $\displaystyle\int\sqrt{x^2+9}\,dx$.

$\dfrac{dx}{du}=3\cosh u$, and $x^2+9=9\sinh^2u+9=9\cosh^2u$, so $\sqrt{x^2+9}=3\cosh u$ (taking $\cosh u\geq0$).

$$\int\sqrt{x^2+9}\,dx=\int3\cosh u\cdot3\cosh u\,du=9\int\cosh^2u\,du=9\int\frac{1+\cosh2u}2\,du$$

$$=\frac92\left(u+\frac{\sinh2u}2\right)+c=\frac92u+\frac94\times2\sinh u\cosh u+c=\frac92u+\frac92\sinh u\cosh u+c$$

Back-substitute $u=\operatorname{arsinh}\dfrac x3$, $\sinh u=\dfrac x3$, $\cosh u=\dfrac{\sqrt{x^2+9}}3$:

$$\int\sqrt{x^2+9}\,dx=\frac92\operatorname{arsinh}\frac x3+\frac x2\sqrt{x^2+9}+c$$

Exam-style questions

Question 1 4 marks

Differentiate with respect to $x$: (a) $\cosh^33x$, (b) $\ln(\sinh x)$.

(a) $\dfrac{d}{dx}\left(\cosh^33x\right)=3\cosh^23x\times3\sinh3x=9\sinh3x\cosh^23x$. (M1, A1)

(b) $\dfrac{d}{dx}\left(\ln(\sinh x)\right)=\dfrac{\cosh x}{\sinh x}=\coth x$. (M1, A1)

Question 2 7 marks

Using the substitution $x=2\cosh u$, find $\displaystyle\int\frac1{\sqrt{x^2-4}}\,dx$, for $x>2$, giving your answer in terms of $\operatorname{arcosh}$.

$\dfrac{dx}{du}=2\sinh u$, and $x^2-4=4\cosh^2u-4=4\sinh^2u$, so $\sqrt{x^2-4}=2\sinh u$. (M1, A1)

$$\int\frac1{\sqrt{x^2-4}}\,dx=\int\frac{2\sinh u}{2\sinh u}\,du=\int1\,du=u+c$$ (M1, A1)

Back-substitute $u=\operatorname{arcosh}\dfrac x2$: $\displaystyle\int\frac1{\sqrt{x^2-4}}\,dx=\operatorname{arcosh}\frac x2+c$. (A1, A1, A1)

Question 3 6 marks

Show that $\displaystyle\int_0^{\ln2}\tanh x\,dx=\ln\frac54$.

$\displaystyle\int\tanh x\,dx=\int\frac{\sinh x}{\cosh x}\,dx=\ln(\cosh x)+c$ (since $\cosh x>0$ always). (M1, A1)

$\displaystyle\int_0^{\ln2}\tanh x\,dx=\Big[\ln(\cosh x)\Big]_0^{\ln2}=\ln(\cosh(\ln2))-\ln(\cosh0)$. (M1)

$\cosh(\ln2)=\dfrac{e^{\ln2}+e^{-\ln2}}2=\dfrac{2+\frac12}2=\dfrac54$, and $\cosh0=1$. (A1, A1)

So the integral is $\ln\dfrac54-\ln1=\ln\dfrac54$, as required. (A1)

← Previous

Identities & Equations

Differential Equations: First-Order