Core Pure · Hyperbolic Functions · Year 2

Differentiating Hyperbolic Functions

Standard derivatives of sinh, cosh, tanh and their inverses, and why Osborn's rule doesn't apply here.

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

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 6 marks

Find the exact coordinates of the stationary point of $y=\cosh2x-4\sinh x$.

$\dfrac{dy}{dx}=2\sinh2x-4\cosh x=4\sinh x\cosh x-4\cosh x=4\cosh x(\sinh x-1)$, using $\sinh2x=2\sinh x\cosh x$. (M1, A1, M1)

Since $\cosh x>0$ for all $x$, stationary points occur where $\sinh x=1$, i.e. $x=\operatorname{arsinh}1=\ln\left(1+\sqrt2\right)$. (M1, A1)

At this point, $\cosh2x=1+2\sinh^2x=1+2=3$, so $y=3-4(1)=-1$. Stationary point: $\left(\ln\left(1+\sqrt2\right),-1\right)$. (A1)

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Identities & Equations

Integrating Hyperbolic Functions