If a function $f$ has an inverse $f^{-1}$, the derivative of the inverse is related to the derivative of the original function. This relationship is elegant: the derivative of the inverse at a point is the reciprocal of the derivative of the original function at the corresponding point.
🎯 In this section you will learn
- The formula for the derivative of an inverse function: $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$
- Derivatives of $\arcsin x$, $\arccos x$, and $\arctan x$
- How to derive these formulas (and why they make sense)
📌 The Inverse Function Derivative Formula
If $f$ is one-to-one and differentiable, and $f'(f^{-1}(a)) \neq 0$, then:
$$ (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))} $$
Equivalently, if $y = f^{-1}(x)$, then $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.
💡 Intuition
The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. Slopes are reciprocals at corresponding points.
Let $f(x) = x^3 + 2$, find $(f^{-1})'(10)$.
①
Find $f^{-1}(10)$
Solve $x^3 + 2 = 10$ ⇒ $x^3 = 8$ ⇒ $x = 2$. So $f^{-1}(10) = 2$.
②
Find $f'(x)$
$f'(x) = 3x^2$
③
Apply formula
$(f^{-1})'(10) = \frac{1}{f'(2)} = \frac{1}{3(2)^2} = \frac{1}{12}$
📌 Derivatives of Inverse Trigonometric Functions
The inverse trig functions — $\arcsin x$, $\arccos x$, $\arctan x$ — are extremely important in calculus. Their derivatives come from the inverse function formula (or implicit differentiation).
| Function | Derivative | Domain (for derivative) |
|---|
| $\frac{d}{dx}\arcsin x$ | $\frac{1}{\sqrt{1-x^2}}$ | $|x| < 1$ |
| $\frac{d}{dx}\arccos x$ | $-\frac{1}{\sqrt{1-x^2}}$ | $|x| < 1$ |
| $\frac{d}{dx}\arctan x$ | $\frac{1}{1+x^2}$ | All real $x$ |
💡 Memory Trick
- Derivative of $\arcsin x$ and $\arccos x$ are opposites (one positive, one negative).
- Derivative of $\arctan x$ looks like the derivative of $\tan x$ ($\sec^2 x$) but reciprocal.
Find $\frac{d}{dx}\arcsin x$ using implicit differentiation.
①
Let $y = \arcsin x$
Then $\sin y = x$, with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.
②
Differentiate implicitly
$\cos y \cdot \frac{dy}{dx} = 1$
③
Solve for $\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\cos y}$
④
Use $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$
Since $\cos y \ge 0$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$
Find $\frac{d}{dx}\arctan x$.
- Let $y = \arctan x$, then $\tan y = x$
- Differentiate: $\sec^2 y \cdot \frac{dy}{dx} = 1$
- $\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$
Find $\frac{d}{dx}\arcsin(2x)$.
- $\frac{d}{dx}\arcsin(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
- Here $u = 2x$, $du/dx = 2$
- $\frac{d}{dx}\arcsin(2x) = \frac{1}{\sqrt{1 - (2x)^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4x^2}}$
Find $\frac{d}{dx}\arctan(x^2)$.
- $\frac{d}{dx}\arctan(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$
- Here $u = x^2$, $du/dx = 2x$
- $\frac{d}{dx}\arctan(x^2) = \frac{1}{1 + (x^2)^2} \cdot 2x = \frac{2x}{1 + x^4}$
Find $\frac{d}{dx}\arccos(3x)$.
- $\frac{d}{dx}\arccos(u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
- Here $u = 3x$, $du/dx = 3$
- $\frac{d}{dx}\arccos(3x) = -\frac{1}{\sqrt{1 - 9x^2}} \cdot 3 = -\frac{3}{\sqrt{1 - 9x^2}}$
⚠️ Common Mistakes
- Forgetting the domain restrictions: $\arcsin x$ and $\arccos x$ are only defined for $|x| \le 1$, and their derivatives require $|x| < 1$.
- Sign errors with $\arccos$: $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$, not positive!
- Confusing $\arctan x$ with $\cot x$: They are different functions with different derivatives.
- Forgetting the Chain Rule: For $\arcsin(2x)$, don't forget the $2$ from the derivative of $2x$.
🔍 Key Takeaways
- Inverse function derivative formula: $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$.
- Memorize the three inverse trig derivatives: $\arcsin x$, $\arccos x$, $\arctan x$.
- $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$, $|x| < 1$.
- $\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$, $|x| < 1$.
- $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$, all real $x$.
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