# 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?

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LaTeX tip: There are preexisting operators \sin and \cos for $\sin x$ and $\cos x$. There is a quick reference guide here. – Aeolian Mar 6 '13 at 7:50

\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.

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.$$\frac{d}{dx} f(x) = \frac{d}{dx} q(\pi x) \times \frac{d}{dx} \pi x$$
Then, shouldn't be like this? $$f'(x)={\pi}cos$$ why is ${\pi}x$ in the answer again. thx – Sb Sangpi Mar 6 '13 at 8:01
@SbSangpi Because the chain rule indicates that if $$f(x)=g(h(x))$$ then $$f'(x) = g'(h(x))h'(x)$$ $\pi x$ is the $h(x)$. – Archy Wilhes Mar 6 '13 at 8:06
@SbSangpi As a function, cosine needs an input, in other words $x$, or rather in this case $\pi x$. On its own, "$\cos$", is meaningless. – Jp McCarthy Mar 6 '13 at 9:06