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Wednesday, April 16, 2014

Implicit differentiation: a motivating example

   Without getting into technical definitions of what is an implicit function and when a given equation in two variables defines implicitly one or more -differentiable or not- functions (i will leave such a discussion for some subsequent post on theory), I will just examine a simple yet illuminating example:
   Let us consider the function $y=f(x)= \frac{1}{x} \equiv x^{-1}$. It is well known that the power rule of differentiation $x^{a} = ax^{a-1}$ applies for any real value of $a$ (in its respective domain of course). So we can readily conclude that $f'(x) = - \frac{1}{x^{2}}$ in $\mathbb{R}^{*}$. 
   But let us momentarily think a little different: Since $y=\frac{1}{x} \Rightarrow xy=1$ we have that  $$xf(x)=1 \Leftrightarrow xy=1$$. 
We differentiate this last relation, applying the product rule of differentiation to both sides and we get $$f(x)+xf'(x)=0 \Leftrightarrow y+xy'=0$$, thus $y'=f'(x)=-\frac{f(x)}{x}=-\frac{y}{x}$. Using the definition $y=f(x)= \frac{1}{x}$ we finally arrive at $$y'=f'(x) = - \frac{1}{x^{2}}$$ 
in $\mathbb{R}^{*}$.

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