Differentiation of Inverse Functions

Understanding Inverse Functions

An inverse function, denoted as $f^{-1}(x)$ (read as "f inverse of x"), essentially "undoes" or "reverses" the action of the original function $f(x)$. If applying $f$ to an input $a$ yields an output $b$ (i.e., $f(a) = b$), then applying the inverse function $f^{-1}$ to $b$ will return the original input $a$ (i.e., $f^{-1}(b) = a$).

For a function to possess an inverse that is itself a function, the original function must be one-to-one (or injective). This means that each distinct input value maps to a unique output value; no two different inputs produce the same output. Graphically, a function is one-to-one if it passes the horizontal line test (any horizontal line intersects the graph at most once).

Many common functions, such as $y = x^2$ or trigonometric functions like $y = \sin x$, are not one-to-one over their entire natural domains. To define their inverse functions, we restrict their domains to specific intervals where they *are* one-to-one. For example:

• $y = \sin x$ is restricted to the domain $[-\pi/2, \pi/2]$ (range $[-1, 1]$) to define $y = \arcsin x$ (which has domain $[-1, 1]$ and range $[-\pi/2, \pi/2]$).
• $y = \cos x$ is restricted to $[0, \pi]$ (range $[-1, 1]$) for $y = \arccos x$ (domain $[-1, 1]$, range $[0, \pi]$).
• $y = \tan x$ is restricted to $(-\pi/2, \pi/2)$ (range $(-\infty, \infty)$) for $y = \arctan x$ (domain $(-\infty, \infty)$, range $(-\pi/2, \pi/2)$).

Geometrically, the graph of an inverse function $y=f^{-1}(x)$ is a reflection of the graph of $y=f(x)$ across the line $y=x$. This symmetry is a key visual characteristic of inverse functions.

If a function $f$ is differentiable and has an inverse $f^{-1}$, and importantly, if $f'(f^{-1}(x)) \neq 0$ (meaning the tangent line to $f$ at the point $(f^{-1}(x), x)$ is not horizontal), then the derivative of the inverse function can be found using the formula:

$$ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} $$

This formula states that the derivative of the inverse function at a point $x$ is the reciprocal of the derivative of the original function evaluated at $f^{-1}(x)$.

Deriving Derivatives of Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions can be derived using implicit differentiation, starting from the definition of the inverse function.

Derivative of $y = \arcsin x$

1. Start with the definition: If $y = \arcsin x$, this implies $\sin y = x$. We must also consider the restricted range of $y$, which is $[-\pi/2, \pi/2]$, to ensure $\arcsin x$ is a function.
2. Differentiate implicitly: Differentiate both sides of $\sin y = x$ with respect to $x$. We use the chain rule on the left side because $y$ is a function of $x$:

   $\frac{d}{dx}(\sin y) = \frac{d}{dx}(x)$

   $\cos y \cdot \frac{dy}{dx} = 1$

   This step applies the chain rule: the derivative of $\sin y$ is $\cos y$, and then we multiply by the derivative of the inner function $y$ with respect to $x$, which is $\frac{dy}{dx}$.
3. Solve for $\frac{dy}{dx}$: Algebraically isolate $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{\cos y}$

4. Express $\cos y$ in terms of $x$: Our derivative should be in terms of $x$, not $y$. We use the Pythagorean identity $\sin^2 y + \cos^2 y = 1$. This gives $\cos^2 y = 1 - \sin^2 y$. Since $y \in [-\pi/2, \pi/2]$, $\cos y$ is non-negative (i.e., $\cos y \ge 0$). Therefore, we take the positive square root: $\cos y = \sqrt{1 - \sin^2 y}$.
5. Substitute $\sin y = x$: From our initial definition, $\sin y = x$. Substitute this into the expression for $\cos y$:

   $\cos y = \sqrt{1 - x^2}$

6. Final derivative: Substitute this back into the expression for $\frac{dy}{dx}$:

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

   This derivative is valid for $|x| < 1$, as the denominator would be zero or undefined at $x = \pm 1$.

Similar procedures are used to find the derivatives of $\arccos x$, $\arctan x$, etc., each utilizing the appropriate trigonometric identities and considering the restricted ranges of the inverse functions.

Summary: Derivatives of Inverse Trigonometric Functions

These are the standard derivative rules for inverse trigonometric functions, incorporating the chain rule where $u$ is a differentiable function of $x$. Domain restrictions are important for these derivatives to be valid.

• $\frac{d}{dx}(\arcsin u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$, for $|u| < 1$.
• $\frac{d}{dx}(\arccos u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$, for $|u| < 1$.
• $\frac{d}{dx}(\arctan u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}$, for all $u$.
• $\frac{d}{dx}(\text{arccot } u) = -\frac{1}{1 + u^2} \cdot \frac{du}{dx}$, for all $u$.
• $\frac{d}{dx}(\text{arcsec } u) = \frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}$, for $|u| > 1$.
• $\frac{d}{dx}(\text{arccsc } u) = -\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx}$, for $|u| > 1$.

Note on $\text{arcsec } u$ and $\text{arccsc } u$: Some textbooks or resources may define these inverse functions with different ranges, which can lead to variations in their derivatives (specifically, the presence or absence of the absolute value $|u|$ in the denominator). The forms presented here are common and consistent with many calculus texts and software like Math.js.

Inverse Hyperbolic Functions: Logarithmic Forms

Inverse hyperbolic functions, like their trigonometric counterparts, have specific definitions and can also be expressed using natural logarithms. These logarithmic forms are particularly useful for deriving their derivatives and for computations.

• $\text{arsinh } x = \ln(x + \sqrt{x^2 + 1})$ (defined for all real $x$)
• $\text{arcosh } x = \ln(x + \sqrt{x^2 - 1})$ (defined for $x \ge 1$)
• $\text{artanh } x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$ (defined for $|x| < 1$)
• $\text{arccsch } x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)$ (defined for $x \neq 0$)
• $\text{arsech } x = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ (defined for $0 < x \le 1$)
• $\text{arcoth } x = \frac{1}{2} \ln\left(\frac{x+1}{x-1}\right)$ (defined for $|x| > 1$)

These logarithmic representations allow us to use the known rules for differentiating logarithms and algebraic functions to find the derivatives of inverse hyperbolic functions.

Deriving Derivatives of Inverse Hyperbolic Functions

The derivatives of inverse hyperbolic functions are typically found by differentiating their logarithmic forms using the chain rule.

Derivative of $y = \text{arsinh } x$

1. Start with the logarithmic form: $y = \ln(x + \sqrt{x^2+1})$. This is the definition we'll differentiate.
2. Apply the chain rule for logarithms: Let $u = x + \sqrt{x^2+1}$. Then $y = \ln u$, and $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$.
3. Find $\frac{du}{dx}$: We need to differentiate $u = x + \sqrt{x^2+1}$ with respect to $x$:

   $\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}((x^2+1)^{1/2})$

   $\frac{du}{dx} = 1 + \frac{1}{2}(x^2+1)^{-1/2} \cdot (2x) = 1 + \frac{x}{\sqrt{x^2+1}}$

   To combine these terms, find a common denominator:

   $\frac{du}{dx} = \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} + \frac{x}{\sqrt{x^2+1}} = \frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}$

4. Substitute back into $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}$

5. Simplify: The term $(x+\sqrt{x^2+1})$ cancels out:

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

A similar process involving differentiation of the logarithmic forms and application of the chain rule is used to derive the derivatives for other inverse hyperbolic functions like $\text{arcosh } x$ and $\text{artanh } x$.

Summary: Derivatives of Inverse Hyperbolic Functions

If $u$ is a differentiable function of $x$, the standard derivative rules for inverse hyperbolic functions (again, incorporating the chain rule) are as follows, with their respective domain considerations:

• $\frac{d}{dx}(\text{arsinh } u) = \frac{1}{\sqrt{u^2+1}} \cdot \frac{du}{dx}$ (for all $u$)
• $\frac{d}{dx}(\text{arcosh } u) = \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx}$ (for $u > 1$)
• $\frac{d}{dx}(\text{artanh } u) = \frac{1}{1-u^2} \cdot \frac{du}{dx}$ (for $|u| < 1$)
• $\frac{d}{dx}(\text{arccsch } u) = -\frac{1}{|u|\sqrt{1+u^2}} \cdot \frac{du}{dx}$ (for $u \neq 0$)
• $\frac{d}{dx}(\text{arsech } u) = -\frac{1}{u\sqrt{1-u^2}} \cdot \frac{du}{dx}$ (for $0 < u < 1$)
• $\frac{d}{dx}(\text{arcoth } u) = \frac{1}{1-u^2} \cdot \frac{du}{dx}$ (for $|u| > 1$)

Notice that the derivative of $\text{artanh } u$ and $\text{arcoth } u$ have the same algebraic form, $\frac{1}{1-u^2}$, but they are valid over different domains, which is critical in applications.