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

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.

If $f(x)$ is increasing, $f'(x) > 0$, so $(f^{-1})'(x) > 0$ as well, meaning the graph of $(f^{-1})'(x)$ is positive.

The graph of the inverse function is the reflection of the original function across the line $y = x$.

A vertical tangent line on $f(x)$ means $f'(x) = 0$ at that point, which implies the derivative of the inverse function is undefined (has a vertical asymptote) at the corresponding point.

The concavity of $f(x)$ affects the rate of change of $f'(x)$, which in turn affects the shape of $(f^{-1})'(x)$. A concave up $f(x)$ may lead to a different shape for $(f^{-1})'(x)$ compared to a concave down $f(x)$.

It means $f(x)$ is an odd function, i.e., $f(-x) = -f(x)$.

If $f'(x) > 0$, then $f(x)$ is increasing, and $f^{-1}(x)$ is also increasing. If $f'(x) < 0$, then $f(x)$ is decreasing, and $f^{-1}(x)$ is also decreasing.

A sharp corner in $f(x)$ means it's not differentiable at that point, which can affect the differentiability of $f^{-1}(x)$ at the corresponding point.

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

A horizontal asymptote in $f(x)$ becomes a vertical asymptote in $f^{-1}(x)$.

If $f(x)$ is linear, $f'(x)$ is constant, so $(f^{-1})'(x)$ is also constant.