Math Problem Solver

The first few derivatives of \(y = \left(1 + 2\right) \cdot \cos\left( \left(2 - 5\right) \cdot t + 3 \right)\) with respect to \(t\) are: \[ \begin{aligned} \frac{d y}{d t} &= 9 \sin{\left(3 - 3 t \right)}\\ \frac{d^{2} y}{d t^{2}} &= - 27 \cos{\left(3 - 3 t \right)}\\ \frac{d^{3} y}{d t^{3}} &= - 81 \sin{\left(3 - 3 t \right)}\\ \frac{d^{4} y}{d t^{4}} &= 243 \cos{\left(3 - 3 t \right)}\\ \frac{d^{5} y}{d t^{5}} &= 729 \sin{\left(3 - 3 t \right)} \end{aligned} \]

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

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