Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 10E (Analysis)

The function $f$ is differentiable at $a$ when there is a real number $L$ such thatEquivalently, the difference quotient has a finite limit, and

WriteDivide by $h$ and let $h\to 0$. Differentiability implies continuity, so $g(a+h)\to g(a)$. Therefore the product rule gives

No. Take $a=0$ andThen $f$ is differentiable andSince $x|x|/x=|x|\to 0$, the product is differentiable at zero withHowever, $g$ has one-sided derivatives $-1$ and $1$ and is not differentiable at zero. This is a nondifferentiable factor with a differentiable nondegenerate product.

Differentiability lets us writeand, with $k\to 0$,Substitute $k=f(a+h)-f(a)=f^{\prime }(a)h+r(h)$. Since $k=O(h)$, the remainder $s(k)$ is $o(h)$. Thusand the chain rule is

No. At $a=0$, letThe function $g$ is differentiable at $f(0)=0$, andso the composite has derivative one at zero. Butis unbounded, so $f$ is not differentiable. This is a nondifferentiable inner function with a differentiable nondegenerate composite.

To say that $f$ is twice differentiable at $a$ means that $f^{\prime }$ is defined in a neighbourhood of $a$ and is differentiable at $a$; its derivative is $f^{\prime \prime }(a)$.

By the chain rule, near $a$,Since $f$ is differentiable and $g^{\prime }$ is differentiable at $f(a)$, another use of the chain rule shows that $g^{\prime }\circ f$ is differentiable at $a$. The product rule then proves that the displayed product is differentiable. Hence $g\circ f$ is twice differentiable, with the second derivative chain rule