Using the quotient rule, the answer is \frac{d}{dx}((sin(x))/x)=\frac{xcos(x)-sin(x)}{x^{2}}ddx(sin(x)x)=xcos(x)−sin(x)x2
While this is technically only true for x!=0x≠0, an interesting thing about this example is that its discontinuity and lack of differentiability at x=0x=0 can be "removed".
Let f(x)=sin(x)/xf(x)=sin(x)x. Use your calculator to graph this over some window near x=0x=0. You'll see that the graph has no vertical asymptote, in spite of the fact that the function is undefined at x=0x=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.
