Suppose f is differentiable at x, but not continuous at x. To violate continuity, there must be some ε, such that f(x+h) is not within ε of f(x), for arbitrarily small values of h. Recall that the difference quotient is f(x+h)-f(x) over h. The numerator doesn't settle down as the denominator goes to 0, hence the fraction jumps towards infinity. It certainly doesn't approach a fixed limit, hence f is not differentiable at x.
A continuous function need not be differentiable, as shown by abs(x) at x = 0. The right derivative is 1, while the left derivative is -1, hence there is no common derivative.
Next consider the right derivative of the square root of x at x = 0. The difference quotient becomes sqrt(h)/h, or 1/sqrt(h), which approaches infinity as h approaches 0. However, the fraction doesn't jump around; it steadily increases. Therefore, some say this function is differentiable at 0, with a derivative of positive infinity. In other words, there is a line tangent to the curve at the origin, and it runs straight up the y axis, with infinite slope. Still, most say the function is not differentiable unless the derivative is a real number. Many theorems, such as the product rule, require a finite derivative.