Differentiating sinh x and cosh x

The derivative of sinh

Differentiate the definition, $\sinh x=\frac{e^x-e^{-x}}{2}$sinh x = (eˣ − e⁻ˣ)/2, one term at a time. The derivative of $e^x$eˣ is $e^x$eˣ. For $e^{-x}$e⁻ˣ the chain rule brings out the derivative of −x, which is −1, so the derivative of $e^{-x}$e⁻ˣ is $-e^{-x}$−e⁻ˣ, and the derivative of $-e^{-x}$−e⁻ˣ is $+e^{-x}$+e⁻ˣ.

So $\frac{d}{dx}\sinh x=\frac{e^x+e^{-x}}{2}$d/dx sinh x = (eˣ + e⁻ˣ)/2, which is cosh x.

Check with a difference quotient at x = 1. $\frac{\sinh 1.01-\sinh 1}{0.01}=1.548982$(sinh 1.01 − sinh 1)/0.01 = 1.548982, and with a step of 0.001 the quotient is 1.543668. They close in on cosh 1 = 1.543081.

cosh x is never less than 1, so the gradient of sinh x is always at least 1. That is why sinh x increases everywhere. At the origin the gradient is cosh 0 = 1, so the tangent there is y = x.

The derivative of cosh

Differentiate $\cosh x=\frac{e^x+e^{-x}}{2}$cosh x = (eˣ + e⁻ˣ)/2 the same way. $e^x$eˣ gives $e^x$eˣ and $e^{-x}$e⁻ˣ gives $-e^{-x}$−e⁻ˣ, so $\frac{d}{dx}\cosh x=\frac{e^x-e^{-x}}{2}$d/dx cosh x = (eˣ − e⁻ˣ)/2, which is sinh x.

Unlike cosine, no minus sign appears. The derivative of cos x is −sin x, but the derivative of cosh x is +sinh x. Check at x = 1: $\frac{\cosh 1.01-\cosh 1}{0.01}=1.182936$(cosh 1.01 − cosh 1)/0.01 = 1.182936, and with a step of 0.001 it is 1.175973. They close in on sinh 1 = 1.175201, which is positive.

The gradient of cosh can still be negative. At x = −1 it is sinh(−1) = −1.175201, because cosh falls as x approaches 0 from the left. The minus there comes from the value of sinh x, not from the rule. At x = 0 the gradient is sinh 0 = 0, at the lowest point of the curve.

Differentiate twice and each function comes back with its own sign: the second derivative of cosh x is cosh x, and the second derivative of sinh x is sinh x. For cos x and sin x, two differentiations change the sign.

The gold curve is y = cosh x and the plain curve is y = sinh x. The gold straight line is the tangent to cosh at x = 1, and its gradient, 1.175201, is the height of the sinh curve at x = 1, the lower marked point. The tangent crosses the y-axis at $e^{-1}=0.367879$e⁻¹ = 0.367879. Left of 0, sinh x is negative and cosh falls.

The derivative of tanh

$\tanh x=\frac{\sinh x}{\cosh x}$tanh x = (sinh x)/(cosh x), so use the quotient rule: the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. That is $\frac{d}{dx}\tanh x=\frac{\cosh x\times \cosh x-\sinh x\times \sinh x}{{\cosh }^{2}x}$d/dx tanh x = (cosh x × cosh x − sinh x × sinh x)/cosh²x.

The numerator is ${\cosh }^{2}x-{\sinh }^{2}x$cosh²x − sinh²x, which the identity says is 1. So $\frac{d}{dx}\tanh x=\frac{1}{{\cosh }^{2}x}$d/dx tanh x = 1/cosh²x, which is also written sech²x.

At x = 0 the gradient is $\frac{1}{{\cosh }^{2}0}=1$1/cosh²0 = 1. At x = 0.5 it is 0.786448, and at x = 1 it is 0.419974. It is positive everywhere, so tanh x always increases, and it shrinks toward 0 far from the origin, where tanh x levels off toward 1 and −1.

The gold curve is y = tanh x and the plain one is its gradient, $y=\frac{1}{{\cosh }^{2}x}$y = 1/cosh²x. The gradient is highest, 1, at x = 0, where tanh is steepest, and it is 0.419974 at x = 1. It never drops to 0, and tanh never stops rising.

The integrals

Read each derivative backwards. sinh x differentiates to cosh x, so $\int \cosh xdx=\sinh x+c$∫ cosh x dx = sinh x + c. cosh x differentiates to sinh x, so $\int \sinh xdx=\cosh x+c$∫ sinh x dx = cosh x + c. Again no minus sign appears, where $\int \sin xdx=-\cos x+c$∫ sin x dx = −cos x + c has one.

For example, the area under y = cosh x from x = 0 to x = 1 is sinh 1 − sinh 0 = 1.175201, and the area under y = sinh x over the same stretch is cosh 1 − cosh 0 = 0.543081. Reading the tanh result backwards gives $\int \frac{1}{{\cosh }^{2}x}dx=\tanh x+c$∫ 1/cosh²x dx = tanh x + c.

With the chain rule

A hanging cable has the shape $y=a\cosh \left(\frac{x}{a}\right)$y = a cosh(x/a). The chain rule gives $\frac{d}{dx}\cosh \left(\frac{x}{a}\right)=\left(\frac{1}{a}\right)\sinh \left(\frac{x}{a}\right)$d/dx cosh(x/a) = (1/a) sinh(x/a), so the gradient of the cable is $a\times \left(\frac{1}{a}\right)\sinh \left(\frac{x}{a}\right)=\sinh \left(\frac{x}{a}\right)$a × (1/a) sinh(x/a) = sinh(x/a): the a in front cancels the $\frac{1}{a}$1/a from inside.

Backwards, $\int \cosh \left(\frac{x}{a}\right)dx=a\sinh \left(\frac{x}{a}\right)+c$∫ cosh(x/a) dx = a sinh(x/a) + c, and the a appears in front, because differentiating $a\sinh \left(\frac{x}{a}\right)$a sinh(x/a) gives back $\cosh \left(\frac{x}{a}\right)$cosh(x/a).

The usual mistakes

Copying the minus sign from cosine. $\frac{d}{dx}\cosh x=\sinh x$d/dx cosh x = sinh x, and $\int \sinh xdx=\cosh x+c$∫ sinh x dx = cosh x + c, with no minus sign in either.

Forgetting the chain rule. $36\cosh \left(\frac{x}{36}\right)$36 cosh(x/36) differentiates to $\sinh \left(\frac{x}{36}\right)$sinh(x/36), not to $36\sinh \left(\frac{x}{36}\right)$36 sinh(x/36).

Leaving off the square. $\frac{d}{dx}\tanh x$d/dx tanh x is $\frac{1}{{\cosh }^{2}x}$1/cosh²x, not $\frac{1}{\cosh x}$1/(cosh x).

Two cables

In the first application below, the length of a hanging chain needs its gradient, $\sinh \left(\frac{x}{20}\right)$sinh(x/20). The identity turns $\sqrt{1+{\sinh }^{2}u}$√(1 + sinh²u) into cosh u, which integrates back to a sinh.

In the second, the gradient of a footbridge cable at its anchor is a sinh, and the angle it makes with the horizontal follows from a slope triangle whose hypotenuse the identity turns into a cosh.