Exponential, logarithmic and hyperbolic functions and their derivatives
Differentiate exponentials and logarithms, then use exponentials to build hyperbolic functions.
Exponentials and logarithms
An exponential function has a fixed positive base and a changing exponent. The natural exponential \(e^x\) is distinguished by being its own derivative. Its inverse is the natural logarithm \(\ln x\).
Rules
Exponential and logarithmic derivatives
\[\frac{d}{dx}e^x=e^x,\qquad\frac{d}{dx}\ln x=\frac1x\quad(x>0).\]For a constant base \(b>0\),
\[\frac{d}{dx}b^x=b^x\ln b.\]The exponential rules hold for every real input, including the constant case \(b=1\).
For a differentiable inner function \(u\), the chain rule gives
\[\frac{d}{dx}e^{u(x)}=e^{u(x)}u^{\prime}(x),\] \[\frac{d}{dx}\ln(u(x))=\frac{u^{\prime}(x)}{u(x)},\qquad u(x)>0.\]Derivations from the natural exponential rule (optional)
Taking \(\frac{d}{dx}e^x=e^x\) as the starting rule, the inverse derivative theorem gives
\[\frac{d}{dx}\ln x=\frac1{e^{\ln x}}=\frac1x,\qquad x>0.\]Rewriting the fixed-base exponential gives
\[\frac{d}{dx}b^x=\frac{d}{dx}e^{x\ln b}=e^{x\ln b}\ln b=b^x\ln b,\]The two composition rules follow directly from the chain rule. This derives the other formulas from the natural exponential rule; it does not claim to prove that starting rule.
The base stays fixed in an exponential derivative; a changing base and exponent will need logarithmic differentiation.
Example 1
An exponential with a changing exponent
For \(P(t)=5e^{-t^2}\),
\[P^{\prime}(t)=5e^{-t^2}(-2t)=-10te^{-t^2}.\]The exponential factor remains; the inner derivative supplies the factor \(-2t\). The function and derivative are defined for all real \(t\).
Example 2
A logarithm and its domain
Let \(L(x)=\ln(4-x^2)\). A positive logarithm argument requires \(-2<x<2\). On this interval,
\[L^{\prime}(x)=\frac{-2x}{4-x^2}.\]The same restriction applies to the derivative; a formula with a nonzero denominator alone does not extend the original domain.
Example 3
A product with a natural logarithm
For \(F(x)=x\ln(1+x^2)\), use the product and chain rules:
\[F^{\prime}(x)=\ln(1+x^2)+\frac{2x^2}{1+x^2}.\]Both the function and derivative are defined for every real input, since \(1+x^2>0\).
Hyperbolic functions
Hyperbolic sine and hyperbolic cosine combine the growing exponential \(e^x\) with the decaying exponential \(e^{-x}\). Their names resemble sine and cosine, but these functions do not repeat periodically.
The identity
\[\cosh^2x-\sinh^2x=1\]explains the name: taking \(x=\cosh t\) and \(y=\sinh t\) places the point on the unit hyperbola \(x^2-y^2=1\). The interactive figure below traces its right-hand branch. The identity follows directly from the definitions:
\[\cosh^2x-\sinh^2x=\frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4}=\frac{4e^xe^{-x}}4=1.\]The coordinates always satisfy the unit-hyperbola identity. At \(t=0\) the point is \((1,0)\); negative and positive values trace the lower and upper parts of the same branch.
Graphs of the hyperbolic functions
Hyperbolic derivatives
The exponential rules above give \(\frac{d}{dx}e^{-x}=-e^{-x}\). Differentiate the definitions term by term to obtain the hyperbolic rules.
Rules
The three main derivatives
\[\frac{d}{dx}\sinh x=\cosh x,\qquad \frac{d}{dx}\cosh x=\sinh x.\] \[\frac{d}{dx}\tanh x=\frac{1}{\cosh^2x}=\operatorname{sech}^2x.\]Here \(\operatorname{sech}x=1/\cosh x\) is hyperbolic secant. These derivatives exist for every real input.
Derivations (optional)
Differentiate the exponential definitions term by term:
\[\frac{d}{dx}\sinh x=\frac{e^x+e^{-x}}2=\cosh x,\qquad \frac{d}{dx}\cosh x=\frac{e^x-e^{-x}}2=\sinh x.\]For hyperbolic tangent, use the quotient rule and the identity established above:
\[\frac{d}{dx}\tanh x=\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=\frac1{\cosh^2x}.\]The important sign difference is that \(\cosh x\) differentiates to positive \(\sinh x\); the circular cosine differentiates to \(-\sin x\).
Hyperbolic examples
Example 4
Keep the inner derivative
For \(p(x)=\cosh(2x^2)\), differentiate the outside, then the inside:
\[p^{\prime}(x)=\sinh(2x^2)\cdot4x.\]There is no minus sign from hyperbolic cosine. The factor \(4x\) comes from the inner function \(2x^2\).
Example 5
Two changing factors
For \(q(x)=x^2\sinh x\), use the product rule:
\[q^{\prime}(x)=2x\sinh x+x^2\cosh x.\]The first term differentiates the polynomial factor; the second differentiates hyperbolic sine.
Example 6
The slope of a hanging cable
An ideal uniform cable hanging under its own weight forms a catenary. For a cable model
\[y(x)=2\cosh(x/2)-2,\]with both coordinates in metres, its slope is
\[y^{\prime}(x)=2\sinh(x/2)\cdot\frac12=\sinh(x/2).\]At the centre, \(y(0)=0\) and \(y^{\prime}(0)=0\): the cable has a horizontal tangent. At \(x=2\), its slope is \(\sinh1\approx1.175\), or about \(1.175\) metres of vertical rise per metre of horizontal change.
Reciprocal hyperbolic functions (optional)
The remaining functions are \(\operatorname{sech}x=1/\cosh x\), \(\operatorname{csch}x=1/\sinh x\), and \(\coth x=\cosh x/\sinh x\). Their derivatives follow from the quotient rule:
\[\frac{d}{dx}\operatorname{sech}x=-\operatorname{sech}x\tanh x.\] \[\frac{d}{dx}\operatorname{csch}x=-\operatorname{csch}x\coth x,\qquad \frac{d}{dx}\coth x=-\operatorname{csch}^2x.\]Hyperbolic secant is defined for every real input; hyperbolic cosecant and cotangent exclude \(0\), where \(\sinh0=0\).