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}
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