9.1 Derivatives of inverse functions

Inverse functions flip inputs and outputs, allowing us to "undo" operations. They're like mathematical mirrors, reflecting functions across y=x. Understanding their derivatives is crucial for solving complex problems in calculus.

Knowing how to find derivatives of inverse functions opens doors to tackling optimization and related rates problems. It's a powerful tool that lets us analyze relationships between variables in reverse, expanding our problem-solving toolkit.

Inverse Functions and Their Derivatives

Concept of inverse functions

• Definition states that if $f(x)$ and $g(x)$ are inverse functions, then composing them in either order yields the original input ($f(g(x)) = g(f(x)) = x$)
• Graphically, the inverse function is a reflection of the original function across the line $y = x$ (swaps $x$ and $y$ coordinates of each point)
• Relationship between derivatives: if $f(x)$ and $g(x)$ are inverses and both differentiable, then $g'(x) = \frac{1}{f'(g(x))}$ (reciprocal of derivative evaluated at inverse)
  • Derived using chain rule and fact that $f(g(x)) = x$ (composing inverses yields identity function)

Formula for inverse derivatives

• If $y = f^{-1}(x)$ is the inverse of $x = f(y)$, then $\frac{dy}{dx} = \frac{1}{f'(y)}$ (derivative of inverse equals reciprocal of derivative of original function)
• Steps to find derivative of inverse:
  1. Given $y = f^{-1}(x)$, write $x = f(y)$ (swap $x$ and $y$ to get original function)
  2. Find $\frac{dx}{dy}$ using derivative of original function (differentiate with respect to $y$)
  3. Solve for $\frac{dy}{dx}$ by taking reciprocal of $\frac{dx}{dy}$ (invert to get derivative of inverse)
  4. Replace $y$ with $f^{-1}(x)$ in resulting expression (substitute inverse function for $y$)

Derivatives of inverse trigonometric functions

• $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}$ (derivative of arcsine)
• $\frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1-x^2}}$ (derivative of arccosine, negative due to decreasing function)
• $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$ (derivative of arctangent)
• $\frac{d}{dx} \arccot(x) = -\frac{1}{1+x^2}$ (derivative of arccotangent, negative due to decreasing function)
• $\frac{d}{dx} \arcsec(x) = \frac{1}{|x|\sqrt{x^2-1}}$ (derivative of arcsecant, absolute value since secant is always positive)
• $\frac{d}{dx} \arccsc(x) = -\frac{1}{|x|\sqrt{x^2-1}}$ (derivative of arccosecant, negative due to decreasing function)
• Domain of inverse function equals range of original function (restricts inputs to ensure one-to-one correspondence)
  • $\arcsin(x)$ has domain $[-1, 1]$ since range of $\sin(x)$ is $[-1, 1]$ (sine outputs between -1 and 1)

Applications of inverse function derivatives

• Optimization problems involve finding maximum or minimum values (critical points where derivative equals zero)
• Related rates problems require applying chain rule to derivatives of inverse functions (rates of change depend on each other)
• Strategies:
  1. Identify inverse function and its derivative (recognize function composition)
  2. Apply chain rule when necessary (multiple variables changing with respect to time or each other)
  3. Use domain restrictions to check validity of solution (ensure inputs are within allowed range)