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.

$f'(x)$ is the derivative of the original function. $(f^{-1})'(x)$ requires using the inverse derivative formula: $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

Chain rule: used for composite functions. Inverse function rule: specifically for derivatives of inverse functions.

Explicit: function is defined as y = f(x). Implicit: function is not explicitly solved for y.

For $f(x)$, directly apply differentiation rules. For $f^{-1}(x)$, use the inverse derivative formula or find $f^{-1}(x)$ explicitly and then differentiate.

For $f(x)$, use implicit differentiation directly. For $f^{-1}(x)$, either find $f^{-1}(x)$ explicitly (if possible) and then differentiate, or use the inverse derivative formula in conjunction with implicit differentiation.

Finding $f^{-1}(x)$ involves switching $x$ and $y$ and solving for $y$. Finding $(f^{-1})'(x)$ involves differentiating the inverse function using the inverse derivative rule.

If $f(a) = b$, then $f'(a)$ is the slope of $f(x)$ at $x=a$, and $(f^{-1})'(b)$ is the slope of $f^{-1}(x)$ at $x=b$. These slopes are reciprocals of each other.

The graph of $f(x)$ shows the function's values, while the graph of $f'(x)$ shows the rate of change of $f(x)$.

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

$f'(a)$ is the derivative of $f(x)$ evaluated at $x=a$. $(f^{-1})'(a)$ is the derivative of the inverse function evaluated at $x=a$, and it requires using the inverse derivative formula.

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