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 $1/x$.

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, $(\sin^{-1}(x))' = \frac{1}{\sqrt{1 - x^2}}$.

If $f$ is differentiable at $a$ and $f'(a) \neq 0$, then $f^{-1}$ is differentiable at $f(a)$ and $(f^{-1})'(f(a)) = \frac{1}{f'(a)}$.

If $f$ is continuous on $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c$ in $[a, b]$ such that $f(c) = k$. This helps establish the existence of an inverse function over an interval.

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

If $y = f(u)$ and $u = g(x)$ are both differentiable, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

If $h(x) = \frac{f(x)}{g(x)}$, then $h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$.

If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.

If $f(x) = c \cdot g(x)$, where $c$ is a constant, then $f'(x) = c \cdot g'(x)$.

If $h(x) = f(x) \pm g(x)$, then $h'(x) = f'(x) \pm g'(x)$.

If $h(x) = f(x)g(x)$, then $h'(x) = f'(x)g(x) + f(x)g'(x)$.