Differentiability of Composite Function - Practice Questions & MCQ

Theorems of Differentiability
Theorem 1
If $f(x)$ and $g(x)$ are both differentiable functions at $x=a$, then the following functions are also differentiable at $x=a$.
(i) $\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})$
(ii) $\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})$
(iii) $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}$, provided $\mathrm{g}(\mathrm{a}) \neq 0$

Theorem 2
If $f(x)$ is differentiable at $x=a$ and $g(x)$ is not differentiable at $x=a$, then $f(x) \pm g(x)$ will not be differentiable at $x=a$.
For example, $\cos (x)+|x|$ is not differentiable at $x=0$, as $\cos (x)$ is differentiable at $x=0$, but $|x|$ is not differentiable at $x=0$.
In other cases, $f(x). g(x)$ and $f(x) / g(x)$ may or may not be differentiable at $x=a$, and hence should be checked using LHD-RHD, continuity or graph.
For example, if $f(x)=0$ (differentiable at $x=0$ ) and $g(x)=|x|$ (non-differentiable at $x=0$ ). Their product is $f(x) \cdot g(x)=0$ which is differentiable. But if $f(x)=2$, $g(x)=|x|$, then $f(x)$. $g(x)=2|x|$ is non-differentiable at $x=0$.
Theorem 3
If $f(x)$ and $g(x)$ both are non-differentiable functions at $x=a$, then the function obtained by algebraic operation of $f(x)$ and $g(x)$ may or may not be differentiable at $x=a$. Hence they should be checked.
For example, Let $f(x)=|x|$, not differentiable at $x=0$ and $g(x)=-|x|$ which is also not differentiable at $x=0$. Their sum $=0$ is differentiable and the difference $=2|x|$ is not differentiable. So there is no definite rule.
Theorem 4
Differentiation of a continuous function may or may not be continuous.