Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 3E (Analysis)

If $f$ is differentiable at $a$ and $g$ at $f(a)$, then $g\circ f$ is differentiable at $a$ andWrite $g(f(a)+u)-g(f(a))=u\{g^{\prime }(f(a))+\epsilon (u)\}$, where $\epsilon (u)\to 0$, and put $u=f(a+h)-f(a)$. Division by $h$ and passage to the limit proves the formula.

No. Take $f(x)=x^2$, $a=0$, and $g(y)=|y|$. The function $g$ is not differentiable at $f(0)=0$, but $g(f(x))=x^2$ is differentiable. Neither function is constant on an interval.

No. Take $f(x)=|x|$, $a=0$, and $g(y)=y^2$. Then $f$ is not differentiable at zero, but $g\circ f=x^2$ is differentiable there.

No. TakeBoth functions are nonconstant on every interval and nondifferentiable at zero, but $g(f(x))=2x$ is differentiable.