Math Problem Solver

The first few derivatives of $y = 2 \sin\left( 3 t - 1 \right)$ with respect to $t$ are: $$\begin{aligned} \frac{d y}{d t} &= 6 \cos{\left(3 t - 1 \right)}\\ \frac{d^{2} y}{d t^{2}} &= - 18 \sin{\left(3 t - 1 \right)}\\ \frac{d^{3} y}{d t^{3}} &= - 54 \cos{\left(3 t - 1 \right)}\\ \frac{d^{4} y}{d t^{4}} &= 162 \sin{\left(3 t - 1 \right)}\\ \frac{d^{5} y}{d t^{5}} &= 486 \cos{\left(3 t - 1 \right)} \end{aligned}$$

Given a positive integer $n$, let $r$ denote the remainder when $n$ is divided by $4$. The $n^{\text{th}}$ derivative of $2 \sin{\left(3 t - 1 \right)}$ with respect to $t$ is given by the formula: $$\frac{d^n}{d t^n} \left[ 2 \sin{\left(3 t - 1 \right)} \right] = \begin{cases} 2 \cdot 3^{n} \sin{\left(3 t - 1 \right)} &\text{ when } r = 0\\ \left(-1\right) 2 \cdot 3^{n} \cos{\left(3 t - 1 \right)} &\text{ when } r = 1\\ 2 \cdot 3^{n} \sin{\left(3 t - 1 \right)} &\text{ when } r = 2\\ \left(-1\right) 2 \cdot 3^{n} \cos{\left(3 t - 1 \right)} &\text{ when } r = 3 \end{cases}$$

In general, the $n^{\text{nth}}$ derivative of $a \sin( b x + c)$ is given by the formula: $$\frac{d^n}{d x^n} \left[ a \sin(b x + c) \right] = \begin{cases} a b^n \sin(b x + c) &\text{ when } r = 0\\ a b^n \cos(b x + c) &\text{ when } r = 1\\ -a b^n \sin(b x + c) &\text{ when } r = 2\\ -a b^n \cos(b x + c) &\text{ when } r = 3 \end{cases}$$ In the case of $2 \sin{\left(3 t - 1 \right)}$, $a = 2$, $b = 3$ and $c = -1$.