How do I prove if the following functions are differentiable at the given value? I have been stumped on the attached question for a while. What would I have to do in order to prove if the functions are differentiable at the given value? I'm only stuck on how to start the problem. 

 A: The accepted answer here (by user Simon Fraser) has a common mistake which I have pointed out in comments.
Lets first understand very clearly the difference between continuity and differentiability of $f$ at a point $a$. To check continuity of $f$ at $a$ you need to look at three things $$\lim_{x\to a^{+}} f(x), \lim_{x\to a^{-}} f(x), f(a) $$ and if all the above three exist and are equal the function $f$ is said to be continuous at $a$.
To check differentiability of $f$ at $a$ we need to look at two things $$\lim_{x\to a^{+}} \frac{f(x) - f(a)} {x-a}, \lim_{x\to a^{-}} \frac{f(x) - f(a)} {x-a} $$ and if the above two limits exist and are equal then the function $f$ is differentiable at $a$ and their common value is the derivative $f'(a) $.
You should evaluate these limits for each of the functions given in your question to check continuity and differentiability of functions involved. 

The technique given in the accepted answer uses the right approach for continuity, but for differentiability it uses wrong approach. Specifically it tries to evaluate the limits $$\lim_{x\to a^{+}} f'(x), \lim_{x\to a^{-}} f'(x) $$ in order to evaluate $f'(a) $. Well these limits are needed not to check the differentiability of $f$ but to check the continuity of derivative $f'$ at $a$ (see the first part of my answer which deals with continuity and replace $f$ by $f'$).
Now here is the surprise! The above technique although wrong, works in most cases and is easier to apply because of the fact that derivatives don't have jump discontinuity. But when you are presented with discontinuous derivatives then the above technique fails.
Consider the function $f$ defined by $$f(x) =x^2\sin(1/x),x>0,f(0)=0,f(x)=-x^2\sin(1/x),x<0$$ This function is differentiable everywhere. However at $0$ the derivative $f'$ is discontinuous. You can check that $f'(0)=0$ but the limits $$\lim_{x\to 0^{+}}f'(x),\lim_{x\to 0^{-}} f'(x) $$ don't exist. Another weird example is the function $g$ defined by $$g(x) =x^2,x\in\mathbb {Q}, g(x) =-x^2,x\in\mathbb {R} \setminus \mathbb {Q} $$ The function $g$ is continuous and differentiable only at $0$. We have $g'(0)=0$ but for any $a$ the limits $$\lim_{x\to a^{+}}g'(x), \lim_{x\to a^{-}} g'(x) $$ don't exist.
Thus one should not fall in this trap and rather work with standard definitions. 
A: By the definition of differentiation, a function $f(x)$ is differentiable at a point $a$ iff $\lim\limits_{h\to 0} \dfrac{f(a+h)-f(a)}{h}=f'(a).$ For each question we need to verify continuity and differentiability.
An important note: differentiability implies continuity.
Also, just to simplify things, I computed the derivatives instead of proving them using the definition. For instance, in my answer to $1,$ I should have written something like $$\begin{align}\lim\limits_{k\to 0^-}\dfrac{h(0+k)-h(0)}{k}=\lim\limits_{k\to 0^-} \dfrac{[6(0+k)^3 + 2(0+k)+5]-5}{k}\end{align}$$
$$\begin{align}= \lim\limits_{k\to 0^-} 6k^2 +2=2\end{align}.$$ This is fairly simple but vital.
For $1,$ the function is continuous at $t=5,$ so we proceed to check differentiability. We have $\lim\limits_{t\to0^-}h'(t)=2$ and $\lim\limits_{t\to0^+}h'(t)=\lim\limits_{t\to 0^+}4\cos(2t)=4\neq 2,$ so the piecewise function $h(t)$ is not differentiable there.
For $2,$ the function is also clearly continuous at $x=0,$ so we proceed to check differentiability. Use the fact that $\sin^2 x+\cos^2x=1\;\forall x$ to get that $\lim\limits_{x\to 0^-}g'(x)=\lim\limits_{x\to0^-}0=0$ and evaluate the other limit as follows: $\lim\limits_{x\to0^+}g'(x) = 2\cdot 0(0+1)^{40}+40\cdot 0^2 (0+1)^{39} = 0.$ Thus, the derivatives of the two functions have the same limit at $x=0$ so the piecewise function $g(x)$ is differentiable there.
For $3,$ we need to verify continuity. Clearly the function is discontinuous at $1$ so it is not differentiable. 
