If g is differentiable at x, and f is differentiable at g(x), the derivative of the composition f(g(x)) is f′(g)×g′. This is known as the chain rule for differentiation.
If g(x+h) equals g(x) for arbitrarily small values of h, then g′ has to be 0. Let's assume g′ is nonzero first.
Start with the difference quotient f(g(x+h))-f(g(x)) over h, and multiply top and bottom by g(x+h)-g(x). Pull g′(x) out of the picture, leaving f(g(x+h))-f(g(x)) over g(x+h)-g(x). Now g is continuous at x, so as h approaches 0, g(x+h) approaches g(x). Our fraction approaches f′(g(x)), and the derivative becomes f′(g)×g′.
Next assume g′(x) = 0. Choose a neighborhood about x, so that the difference quotient leading to f′(x) is bounded below b. In other words, the difference f(x+t)-f(x) never exceeds b×t. Now the input to f is g, which is also differentiable at x. Select h so that g(x+h)-g(x) over h doesn't exceed ε. In other words, g(x+h)-g(x) doesn't exceed h×ε. Now the difference f(g(x+h))-f(g(x)) doesn't exceed b×h×ε. Divide by h, and the limit is bounded by b times ε, for arbitrary ε. Thus the derivative is 0, just as g′ is 0, and that completes the proof.
This theorem is known as the chain rule, in its simplest form. Generalized versions of the chain rule are presented in multivariable calculus.