How to find $\frac{d^{40}y}{dx^{40}}$, when $y= \sin x$?

Question

What approach would be ideal in finding $\frac{d^{40}y}{dx^{40}}$, when $y= \sin x$?

Answer 1

Notice that every fourth derivative of $\sin(x)$ is $\sin(x)$ itself. Therefore we can write this statement: $$\text{If $y=\sin x$, then }\frac{d^{4n}y}{dx^{4n}}=\sin(x) \, \, \text{($n \in \mathbb N$)}$$ $40$ is a multiple of $4$, which means that: $$\color{green}{\frac{d^{40}y}{dx^{40}}=\sin(x)}$$ It really is that simple. Hope I helped!

Answer 2

If $y = \sin x$, then $y^{(1)} = y' = \cos x$, and $y^{(2)} = y'' = -\sin x$, $y^{(4)} = \sin x$, so: $y^{(8)} = \sin x$, and $y^{(40)} = \sin x$, here $y^{(n)} = \dfrac{d^{n}y}{dx^n}.$