Implicit differentiation 2

Derivatives of inverse functions.

  • Let \(f(x)\) be a function. Recall that its inverse is defined in a way that \(y=f^{-1}(x)\) means \(f(y)=x\).
    • Suppose that \(f(x)\) is differentiable on its domain. To find the derivative of its inverse, we can implicit derivate the equation \(f(y)=x\). We get \(f'(y)y'=1\), which shows \[ (f^{-1}(x))'=\frac{1}{f'(y)}. \]
  • We can use this to derivate inverse trigonometric functions.
    • Let \(y=\sin^{-1}(x)\). Then applying the formula, we get \[ (\sin^{-1}(x))'=\frac{1}{\cos(\sin^{-1}(x))}. \]
    • Since \(-\frac{\pi}{2}\le y\le\frac{\pi}{2}\), we have \(\cos y\ge1\), and thus \[ \cos y=\sqrt{1-\sin(y)^2}. \] This gives the following. \[ (\sin^{-1}(x))'=\frac{1}{\sqrt{1-\sin(y)^2}}=\frac{1}{\sqrt{1-x^2}}. \]
    • Exercise. Find the derivative of \(y=\tan(x)\).
    • Exercises. 3.5: 50, 52, 58, 66, 68, 70, 76, 80