1. Find $f^{-1}(a) = b$. 2. Find $f'(x)$. 3. Evaluate $f'(b)$. 4. Calculate $(f^{-1})'(a) = \frac{1}{f'(b)}$.

1. Find $g(a)$. 2. Find $g'(a) = \frac{1}{f'(g(a))}$. 3. Use point-slope form: $y - g(a) = g'(a)(x - a)$.

1. Find $x$ such that $f(x) = a$, so $g(a) = x$. 2. Find $f'(x)$ from the table. 3. Calculate $g'(a) = \frac{1}{f'(x)}$.

1. Verify that $f(b) = a$. 2. Find $f'(x)$. 3. Evaluate $f'(b)$. 4. The slope of the tangent line is $\frac{1}{f'(b)}$. 5. Use point-slope form: $y - b = \frac{1}{f'(b)}(x - a)$.

1. Find $f'(x)$. 2. Express $g'(x)$ as $\frac{1}{f'(g(x))}$. 3. If needed, use implicit differentiation or other techniques to find $g(x)$ or simplify the expression.

Check if the derivative of the original function is non-zero at the corresponding point. If $f'(f^{-1}(a)) \neq 0$, then $f^{-1}(x)$ is differentiable at $x = a$.

1. Find the point on the graph of $f(x)$ where $y = a$. Let this point be $(b, a)$. 2. Estimate the slope of the tangent line to $f(x)$ at $x = b$. This is $f'(b)$. 3. Calculate $(f^{-1})'(a) = \frac{1}{f'(b)}$.

1. Apply the chain rule carefully, remembering that the derivative of the outer function is evaluated at the inner function. 2. Use the inverse derivative rule when differentiating the inverse function. 3. Simplify the expression.

1. Find the first derivative $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$. 2. Differentiate this expression using the chain rule and quotient rule. 3. Simplify the result.

Use the formula for the derivative of an inverse function and the derivatives of trigonometric functions. For example, $(\sin^{-1}(x))' = \frac{1}{\sqrt{1 - x^2}}$.

A function that 'reverses' another function. If $f(a) = b$, then $f^{-1}(b) = a$.

A function is differentiable at a point if its derivative exists at that point.

A function that has an inverse function.

The derivative of a function $f(x)$ is a measure of how $f(x)$ changes as $x$ changes.

A line that touches a curve at a point and has the same slope as the curve at that point.

The equation of a line given a point $(x_1, y_1)$ and slope $m$: $y - y_1 = m(x - x_1)$.

The set of all possible input values (x-values) for which the function is defined.

The set of all possible output values (y-values) of the function.

A function $f(x)$ is strictly increasing if, for any $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

The reciprocal of a number $x$ is $1/x$.

The derivative of the inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point. If $f(a) = b$, then $(f^{-1})'(b) = \frac{1}{f'(a)}$.

The graphs of a function and its inverse are reflections of each other across the line $y = x$.

The derivative of a function at a point represents the slope of the tangent line to the function's graph at that point.

A strictly increasing or decreasing function is guaranteed to be one-to-one, and therefore invertible.

It represents the derivative of the original function evaluated at the inverse function, ensuring the correct corresponding point is used for the reciprocal calculation.

At a sufficiently small scale, a differentiable function can be approximated by its tangent line.

The domain of $f(x)$ is the range of $f^{-1}(x)$, and the range of $f(x)$ is the domain of $f^{-1}(x)$.

The original function must be differentiable at the point corresponding to the inverse function's input for the inverse derivative to exist.

$f(x)$ is the original function, and $f^{-1}(x)$ is its inverse, which 'undoes' the operation of $f(x)$.

$f'(x)$ is the derivative of the original function, and $(f^{-1})'(x)$ is the derivative of its inverse.