3.7 Derivatives of Inverse Functions

Derivatives of Inverse Functions

Inverse function theorem application

If $f(x)$ is differentiable and one-to-one on an interval, then its inverse $f^{-1}(x)$ is also differentiable. The derivative of the inverse is:

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

This formula says: to find the derivative of the inverse at some input $x$, first figure out what the inverse gives you (that's $f^{-1}(x)$), then plug that into the derivative of the original function, and take the reciprocal. The original function must be continuous and strictly monotonic (always increasing or always decreasing) on its domain for this to work.

Steps to apply the theorem:

1. Identify the original function $f(x)$ and confirm it's one-to-one on the relevant interval.
2. Find $f^{-1}(x)$ if possible, or at least identify the point where you need the derivative.
3. Compute $f'(x)$.
4. Plug $f^{-1}(x)$ into $f'$ and take the reciprocal: $\frac{1}{f'(f^{-1}(x))}$.

Example: Suppose $f(x) = x^3 + x$ and you want the derivative of $f^{-1}(x)$ at $x = 2$. You can't easily solve for $f^{-1}$ algebraically, but you can check that $f(1) = 2$, so $f^{-1}(2) = 1$. Since $f'(x) = 3x^2 + 1$, you get:

$\frac{d}{dx}f^{-1}(2) = \frac{1}{f'(1)} = \frac{1}{3(1)^2 + 1} = \frac{1}{4}$

This is exactly the situation where the theorem shines: you don't need an explicit formula for the inverse.

Inverse trigonometric function derivatives

These six derivatives come directly from applying the inverse function theorem to the standard trig functions (restricted to domains where they're one-to-one). You should memorize them:

• $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$
• $\frac{d}{dx}\text{arccot}(x) = -\frac{1}{1+x^2}$
• $\frac{d}{dx}\text{arcsec}(x) = \frac{1}{|x|\sqrt{x^2-1}}$
• $\frac{d}{dx}\text{arccsc}(x) = -\frac{1}{|x|\sqrt{x^2-1}}$

Notice the pattern: arcsin/arccos are negatives of each other, arctan/arccot are negatives of each other, and arcsec/arccsc are negatives of each other. This happens because each pair sums to a constant (e.g., $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$), so their derivatives must be negatives.

Application examples:

• The slope of the tangent line to $y = \arctan(x)$ at $x = 1$:

$\frac{1}{1+1^2} = \frac{1}{2}$

• Suppose an angle is given by $\theta = \arcsin\!\left(\frac{x}{5}\right)$. By the chain rule, the rate of change with respect to $x$ is:

$\frac{d\theta}{dx} = \frac{1}{\sqrt{1-\left(\frac{x}{5}\right)^2}} \cdot \frac{1}{5}$

At $x = 3$: $\frac{1}{\sqrt{1-\frac{9}{25}}} \cdot \frac{1}{5} = \frac{1}{\sqrt{\frac{16}{25}}} \cdot \frac{1}{5} = \frac{1}{\frac{4}{5}} \cdot \frac{1}{5} = \frac{5}{4} \cdot \frac{1}{5} = \frac{1}{4}$

Don't forget the chain rule factor of $\frac{1}{5}$ here. The original guide listed $\frac{5}{4}$ for this problem, but that omits the chain rule.

Slope relationships for inverse functions

Since $f$ and $f^{-1}$ are reflections of each other across the line $y = x$, their tangent line slopes at corresponding points are reciprocals.

Concretely: if the tangent to $f(x)$ at $(a, f(a))$ has slope $m$, then the tangent to $f^{-1}(x)$ at the reflected point $(f(a), a)$ has slope $\frac{1}{m}$.

This follows directly from the inverse function theorem. Evaluating at $x = f(a)$:

$\frac{d}{dx}f^{-1}\big(f(a)\big) = \frac{1}{f'(a)}$

The left side is the slope of $f^{-1}$ at the point $(f(a), a)$. The right side is the reciprocal of the slope of $f$ at $(a, f(a))$. That's the whole relationship.

One thing to watch: if $f'(a) = 0$, the inverse isn't differentiable at that corresponding point (you'd be dividing by zero). Graphically, a horizontal tangent on $f$ becomes a vertical tangent on $f^{-1}$.

Domain, range, and derivative rules

The domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. This swap matters when you're checking where an inverse trig derivative is valid. For instance, $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$ only applies for $-1 < x < 1$ because that's the domain of arcsin (and the denominator is zero at the endpoints).

When differentiating compositions involving inverse functions, you'll almost always need the chain rule alongside the formulas above. For example:

$\frac{d}{dx}\arctan(3x^2) = \frac{1}{1+(3x^2)^2} \cdot 6x = \frac{6x}{1+9x^4}$

The inverse function derivative gives you the "outer" piece, and the chain rule supplies the derivative of the inner function. Forgetting that inner derivative is one of the most common mistakes on exams.