Finding $f'''(x)$ of $f(x)=\sin{\pi}x$ 
Finding $f'''(x)$ of $f(x)=\sin{\pi}x$.

First
$$f'(x)=\frac{d\sin{\pi}x}{d({\pi}x)}\frac{d({\pi}x)}{dx} ={\pi}\cos{\pi}x$$
Second
$$f''(x)=-{\pi^2}\sin{\pi}x$$
Third
$$f'''(x)= -{\pi^3}\cos {\pi}x$$
These are the steps the teacher has written down on the blackboard, I have trouble understanding "How it is solved".
Would you please provide the steps?
 A: $$
\begin{align*}
f(x) = \sin\pi x
\end{align*}
$$
Here, note that $\pi$ is a constant. So treat it like a coefficient of $x$. Progressively, what I will do is to:


*

*Isolate the coefficient since it is not involved in the differentiation

*Apply chain rule by differentiating the trigonometric function, followed by the function "inside" the trigo function - the $\pi x$. Note that $\frac{d}{dx}\begin{pmatrix}\pi x\end{pmatrix} = \pi$.


$$
\begin{align*}
f(x) &= \sin\pi x \\
f'(x) &= (\cos \pi x) \times \pi \space\space\tag{by chain rule}\\
&= \pi \cos \pi x
\\\\\\f''(x) &= \pi (-\sin \pi x)(\pi) \space\space\tag{by chain rule}\\
&= -\pi^2 \sin \pi x
\\\\\\f'''(x) &= \pi^2 (-\sin \pi x)(\pi) \space\space\tag{by chain rule}\\
&= -\pi^3 \cos \pi x
\end{align*}
$$
And that is how you do it.
Comments
I'll add comments here to address any of your future questions:
First, understand that $\cos$ is a trigonometric function. I would recommend you to write $f(x)$ and $f'(x)$ as a composite function. I will do the easier one.

If $p(x)$ = $\pi x$ and $q(x) = \sin x$, then $f(x) = q(p(x))$. To differentiate $f(x)$, apply chain rule.$$\frac{d}{dx} f(x) = \frac{d}{dx} q(\pi x) \times \frac{d}{dx} \pi x$$

