If $f(x)=a_0\cos|x|+a_1\sin|x|+a_2|x|^3$ is differentiable at $x=0$, then
(A) $a_1=0, a_2=0$
(B) $a_0=0, a_1=0$
(C) $a_1=0$
(D) $a_0, a_1, a_2$ can take any real value
The differentiability of $f(x)$ is due to the individual contribution of each of the terms making up $f(x)$ . The graph of $\cos|x|$ does not have any corner. The graph of $\sin|x|$ has a corner at $x=0$. So $a_1$ has to be $0$. Now, my problem starts here. The graph of $|x|^3$ does not have any corner. It is smooth although a little flat near $x=0$. But the graph of it's derivative has a hole at $x=0$. Therefore it is not differentiable at $x=0$. Thus $a_2$ is also $0$. My question is why does the graph of the derivative of $|x|^3$ have a hole at $x=0$ inspite of the fact that the graph of $|x|^3$ has no corners?
To answer your question, let's find the derivative of $|x|^3$.
$f(x)=|x|^3$
$f(x)=x^2|x|$
By the product rule (and by remembering the derivative of $|x|$),
$f'(x)= 2x|x|+\frac{x^3}{|x|}$
We can further simplify:
$f'(x)=\frac{2x|x|^2+x^3}{|x|}$
$f'(x)=\frac{2x^3+x^3}{|x|}$
$f'(x)=\frac{3x^3}{|x|}$
$f'(x)=3x|x|$
Now we can just evaluate this limit to check for continuity.
$\lim_{x\to0}3x|x|=0$
It helps to simplify the derivative; however, if you left it as a fraction, it would be indeterminate at 0-- not undefined.