# If f(x)=sin2x then what is the 75^(th) derivative?

Sep 25, 2017

${f}^{\left(75\right)} \left(x\right) = - {2}^{75} \cos 2 x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 37778931862957161709568 \cos 2 x$

#### Explanation:

If we have:

$f \left(x\right) = \sin 2 x$

Then repeatedly differentiating wrt $x$ we get:

$\setminus \setminus \setminus \setminus f \left(x\right) = \sin 2 x$
$\setminus \setminus \setminus f ' \left(x\right) = 2 \cos 2 x$
$\setminus f ' ' \left(x\right) = 2 \left(- 2 \sin 2 x\right) \setminus \setminus = - {2}^{2} \sin 2 x$
$f ' ' ' \left(x\right) = - {2}^{2} \left(2 \cos 2 x\right) = - {2}^{3} \cos 2 x$
${f}^{\left(4\right)} \left(x\right) = {2}^{4} \sin 2 x$
$\vdots$

After which the pattern continues with a cycle of 4, so we can represent the ${n}^{t h}$ derivative by:

 f^((n))(x) = { (2^nsin2x, n mod 4 = 0), (2^ncos2x, n mod 4 = 1), (-2^nsin2x, n mod 4 = 2), (-2^ncos2x, n mod 4 = 3) :}

Now $75 \mod 4 = 3$, and so we can deduce that:

${f}^{\left(75\right)} \left(x\right) = - {2}^{75} \cos 2 x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 37778931862957161709568 \cos 2 x$