# Question #d0c3c

Sep 27, 2017

-sin x

#### Explanation:

Sin x and cos x work as thus.

If $f \left(x\right) = \sin \left(x\right) ,$ then $\frac{\mathrm{df}}{\mathrm{dx}} = \cos x , \frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = - \sin x , \frac{{d}^{3} f}{\mathrm{dx}} ^ 3 = - \cos \left(x\right) , \frac{{d}^{4} f}{\mathrm{dx}} ^ 4 = \sin x$

In other words, every 4thmderivative is the original function, as is the eighth, the twelfth, etc. (Note that if we have $\sin \left(u \left(x\right)\right) , \frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \cos \left(u \left(x\right)\right)$ instead.

The highest multiple of 4 in our exponent is 84, so we can essentially ignore that part for the derivative.

$\frac{{d}^{86}}{\mathrm{dx}} ^ 86 \sin \left(x\right) = \frac{{d}^{2}}{\mathrm{dx}} ^ 2 \sin \left(x\right) = - \sin \left(x\right)$