What is the derivative of #sinx/x#?

1 Answer
May 30, 2015

Using the quotient rule, the answer is #\frac{d}{dx}((sin(x))/x)=\frac{xcos(x)-sin(x)}{x^{2}}#

While this is technically only true for #x!=0#, an interesting thing about this example is that its discontinuity and lack of differentiability at #x=0# can be "removed".

Let #f(x)=sin(x)/x#. Use your calculator to graph this over some window near #x=0#. You'll see that the graph has no vertical asymptote, in spite of the fact that the function is undefined at #x=0#. In fact, you should see that #lim_{x->0}f(x)=1#.

Therefore, if we declare that we want #f(0)=1#, we will have created a continuous function for all #x# (a piecewise formula should be written to make this definition most clearly).

Moreover, it's also differentiable everywhere and #f'(0)=0#. This is also consistent with the fact that #lim_{x->0}\frac{x\cos(x)-sin(x)}{x^{2}}=0#, as you can check with your calculator.