Differentiating Inverse Functions (DP IB Analysis & Approaches (AA): HL): Revision Note

Differentiating inverse functions

What is meant by an inverse function?

• Some functions are easier to handle with $x$ (rather than $y$) as the subject

  • i.e.  in the form $x=f\left(y\right)$

  • This can be particularly true when dealing with inverse functions

• If $y=f\left(x\right)$ the inverse would be written as $y=f^{-1}\left(x\right)$

  • Finding $f^{-1}\left(x\right)$ can be awkward

  • But a function and its inverse 'cancel' each other

    • This means $y=f^{-1}\left(x\right)\Rightarrow f\left(y\right)=f\left(f^{-1}\left(x\right)\right)\Rightarrow f\left(y\right)=x$

  • So you can write $x=f\left(y\right)$ instead

    • This expresses the same relationship between $x$ and $y$ as $y=f^{-1}\left(x\right)$ does

How do I differentiate inverse functions?

• With $x=f\left(y\right)$ it is easier to differentiate “$x$ with respect to $y$” rather than “$y$ with respect to $x$”

  • I.e. to find $\frac{\mathrm{d}x}{\mathrm{d}y}$ rather than $\frac{\mathrm{d}y}{\mathrm{d}x}$

  • Note that $\frac{\mathrm{d}x}{\mathrm{d}y}$ will be in terms of $y$

• STEP 1
  For the function $y=f\left(x\right)$, the inverse will be $y=f^{-1}\left(x\right)$

  Rewrite $y=f^{-1}\left(x\right)$ as $x=f\left(y\right)$

  • E.g. Find an expression in terms of $y$ for the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of $y=f^{-1}\left(x\right)$, where $f\left(x\right)=x^3-1$

$y=f^{-1}\left(x\right)\Rightarrow x=f\left(y\right)\Rightarrow x=y^3-1$

• STEP 2
  From $x=f\left(y\right)$ find $\frac{\mathrm{d}x}{\mathrm{d}y}$

$\frac{\mathrm{d}x}{\mathrm{d}y}=3y^2$

• STEP 3
  Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ using $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{\displaystyle \frac{\mathrm{d}x}{\mathrm{d}y}}}$  (this will usually be in terms of $y$)

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{\displaystyle \frac{\mathrm{d}x}{\mathrm{d}y}}}=\frac{1}{3y^2}$

• This can be used to find the gradient of the curve $y=f^{-1}\left(x\right)$ at a given point

  • Substitute the $\mathit{y}$-coordinate of the point into your expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$

    • If the $y$-coordinate is not given, you should be able to work it out from the original function and $x$-coordinate

  • E.g. for the function used above, find the gradient of $y=f^{-1}\left(x\right)$ at the point where $x=7$

    • You need to know the value of $y=f^{-1}\left(7\right)$

      • $y=f^{-1}\left(7\right)\Rightarrow f\left(y\right)=7\Rightarrow y^3-1=7\Rightarrow y=2$

    • Substitute $y=f^{-1}\left(7\right)=2$ into the expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{3{\left(2\right)}^2}=\frac{1}{12}$