Hyperbolic and inverse-function derivatives

What you’ll learn

• Differentiate hyperbolic and inverse trigonometric functions.
• Derive inverse-function derivatives using implicit differentiation.

Understand the idea

Differentiate the exponential definitions to see that $\sinh x$ and $\cosh x$ differentiate into each other. Unlike ordinary cosine, differentiating $\cosh x$ introduces no minus sign. Products, quotients and compositions still require the usual rules.

To differentiate an inverse, write an equivalent direct equation and differentiate implicitly. For example, $y=\operatorname{arsinh}x$ gives $x=\sinh y$ and $dy/dx=1/\cosh y=1/\sqrt{1+x^2}$. Domain restrictions and the chosen inverse branch determine the sign of a square root.

Key formulae

$$(\sinh x)'=\cosh x,\quad(\cosh x)'=\sinh x,\quad(\tanh x)'=\operatorname{sech}^2x$$

$$(\arcsin x)'=\frac1{\sqrt{1-x^2}},\qquad(\arctan x)'=\frac1{1+x^2}$$

$$(\operatorname{arsinh}x)'=\frac1{\sqrt{1+x^2}},\quad(\operatorname{arcosh}x)'=\frac1{\sqrt{x^2-1}}\ (x>1)$$

$$(\operatorname{artanh}x)'=\frac1{1-x^2}\quad(|x|<1)$$

Worked example

Differentiate $y=x\sinh(2x)$.

1. Apply the product rule to $x$ and $\sinh(2x)$.
2. The chain rule gives $d(\sinh(2x))/dx=2\cosh(2x)$.
3. Combine the two product-rule terms.

Answer: $y'=\sinh(2x)+2x\cosh(2x)$.