The Derivative of Inverse Functions

If for a function \(y=f(x)\) derivative \(y_x' \neq 0\) then the derivative of inverse function \(x = f^{-1}(x)\) can be written as $$ x_y' = \frac{1}{y_x'} \Rightarrow \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$$ Inital Example - Calculate the \(y_x'\) and \(x_y'\) of function \(y = x + \ln x\). First, let's calculate \(y_x' = \frac{dy}{dx}\) $$y_x' = 1+ \frac{1}{x} = \frac{x+1}{x}$$ The \(x_y' = \frac{dx}{dy}\) is calculated as an inverse of \(\frac{dy}{dx}\) $$x_y' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{\frac{x+1}{x}} = \frac{x}{x+1} $$

• Example 1 Calculate the \(y_x'\) and \(x_y'\) of a function \(y=3x+x^2.\)
  Solution Before determining the derivative of inverse function the derivative of function must be calculated. \begin{eqnarray} y&=& 3x+x^2\\\nonumber y_x' &=& \frac{dy}{dx} = 3 + 3x^2 \\\nonumber y_x' &=& \frac{dy}{dx} = 3(1+x^2)\end{eqnarray} Then \(x_y'\) can be calculated using the formula \(x_y' = \frac{1}{y_x'}\) $$ x_y' = \frac{1}{y_x'} = \frac{1}{3(1+x^2)}$$
• Example 2 Calculate the \(y_x'\) and \(x_y'\) of a function \(y=x-\frac{1}{2}\sin x.\)
  Solution First step is to calculate the derivative of original function \(y_x'.\) \begin{eqnarray}y &=& x - \frac{1}{2}\sin x\\\nonumber y_x'&=& 1-\frac{1}{2}\cos x\end{eqnarray} The derivative of invervese function is determined from \begin{eqnarray}x_y' = \frac{1}{y_x'} = \frac{1}{1-\frac{1}{2}\cos x} \end{eqnarray}
• Example 3 Calculate the \(y_x'\) and \(x_y'\) of a function \(y=0.1x - e^{\frac{x}{2}}.\)
  Solution Detrmine the derivative of original function i.e. \(y_x'.\)\begin{eqnarray}y&=&0.1x + e^{\frac{x}{2}}\\\nonumber y_x' &=& 0.1 + \frac{1}{2}e^{\frac{x}{2}}\end{eqnarray} Now determine the derivative of inverse function using formula \(x_y' = \frac{1}{y_x'}.\)\begin{eqnarray}x_y' &=& \frac{1}{y_x'} = \frac{1}{0.1 + \frac{1}{2}e^{\frac{x}{2}}}\end{eqnarray}