### Rules of differentiation: Rules of computation for the derivative

### The derivative of an inverse function

We recall that the *inverse function* #{f}^{-1}# of the function #f# is the function satisfying \(f(g(y))=y\) for each \(y\) in the *domain* of \(g\) and \(g(f(x))=x\) for each \(x\) in the domain of \(f\). *Differentiation* of the inverse of a function can be carried out as follows.

Inverse function rule for differentiation

The derivative of #{f}^{-1}# is #\dfrac{1}{f'\circ f^{-1}}#, hence, #\left({f}^{-1}\right)'(x) = \dfrac{1}{f'\left(f^{-1}(x)\right)}#.

From the definition of the inverse function follows #f\circ f^{-1}(x)=x#. If we take the derivative to #x# on the left and right hand side, we find:

\[\left(f\circ f^{-1}\right)'(x)=1\]

The left hand side can be expanded using the *chain rule*

\[\begin{array}{rcl}\left(f\circ g\right)'(x)&=&f'\left(g(x)\right)\cdot g'(x)\tiny. \end{array}\]

If we apply this rule with #g=f^{-1}#, we find \[\begin{array}{rcl}\left(f\circ f^{-1}\right)'(x)&=&f'\left(f^{-1}(x)\right)\cdot\left(f^{-1}\right)'(x)\tiny.\end{array}\]

Since this should be equal to #\frac{\dd}{\dd x}x=1#, it follows that\[f'\left(f^{-1}(x)\right)\cdot \left(f^{-1}\right)'(x)=1\tiny.\]

If we divide this equation on the left and right hand side by #f'\left(f^{-1}(x)\right)#, we find \[\left(f^{-1}\right)'(x)=\frac{1}{f'\left(f^{-1}(x)\right)}\tiny.\]

Consider a point #\rv{x_0,y_0}# on the graph of #f(x)#. In #x_0# the graph of #f# increases at a rate of #f'(x_0)#. Then the point is #\rv{y_0,x_0}# on the graph of #f^{-1}(y)#. In #y_0# the graph of #f^{-1}# increases at a rate of #\frac{1}{f'(x_0)}#.

Now that we know the derivative of the inverse function, we can once again look at the derivative of the power function #f(x)=x^a# with #a# being a negative or fractional number.

Power Rule for differentiation

If #a# is a real number unequal to #0#, then the derivative of the function #x^a# at #\ivco{0}{\infty}# is equal to #ax^{a-1}#. In other words, #\frac{\dd}{{\dd}x}x^a = ax^{a-1}#.

We have already proven this in *Derivative of a polynomial function* for the case when #a# is an integer. Let, for the general case, #a = \dfrac{p}{q}# for two natural numbers #p# and #q# which are both unequal to #0# (we can make this choice because the expression is not defined if #q=0# and we have the constant function if #p=0#).

We first calculate the derivative of #x^{\frac{1}{q}}#. This is the case #a = \dfrac{1}{q}#. Because this is the inverse function of #x^q#, we find

\[\frac{{\dd}}{{\dd}x}\left(x^{\frac{1}{q}}\right) =\frac{1}{q(x^ {\frac{1}{q}})^{q-1}} = \frac{1}{q}x^{\frac{1}{q}-1} = ax^{a-1}\tiny.\] Here, the formula is correct. We find the general case with help of the *chain rule*

\[\begin{array}{rcl}\frac{{\dd}}{{\dd}x}\left(x^{\frac{p}{q}}\right)&=&\frac{{\dd}}{{\dd}x}\left(\left(x^{\frac{1}{q}}\right)^p\right)\\ &=&p\left(x^{\frac{1}{q}}\right)^{p-1}\frac{{\dd}}{{\dd}x}\left(x^{\frac{1}{q}}\right)\\&=&p\left(x^{\frac{1}{q}}\right)^{p-1}\frac{1}{q}x^{\frac{1}{q}-1} \\ &=&\frac{p}{q}x^{\frac{p}{q}-\frac{1}{q}+\frac{1}{q}-1} \\ &=& ax^{a-1}\,\tiny,\end{array}\] with which the formula is derived.

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