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

$g'(x) = \frac{1}{f'(g(x))}$

$y - y_1 = m(x - x_1)$

Switch $x$ and $y$ in the equation and solve for $y$.

$f'(x) = 2$

$f(x) = mx + b$

$f^{-1}(x) = \frac{x-b}{a}$

$\frac{d}{dx} [f(g(x))] = f'(g(x)) * g'(x)$

$\frac{d}{dx} x^n = nx^{n-1}$

$\frac{d}{dx} c = 0$

$f'(x)$ 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.

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 $h'(x) = f'(x)g(x) + f(x)g'(x)$.