Find (d³y)/(dx³) for x = e^-tcos t, y = e^-tsin t

Find $\dfrac{d^{3}y}{dx^{3}}$ for $x = e^{-t}\cos t,\ y = e^{-t}\sin t$.

Step 1: Derivatives with respect to $t$

\[ \dot x = -e^{-t}\bigl(\cos t+\sin t\bigr), \qquad \dot y = e^{-t}\bigl(\cos t-\sin t\bigr). \]

Step 2: First derivative

\[ y'_{x} = -\frac{\cos t-\sin t}{\cos t+\sin t} = -\frac{u}{v}, \qquad u = \cos t-\sin t,\ v = \cos t+\sin t. \]

These satisfy $u' = -v$ and $v' = u$, and $u^{2}+v^{2} = 2$.

Step 3: Second derivative

\[ \bigl(y'_{x}\bigr)'_{t} = -\frac{u'v-uv'}{v^{2}} = -\frac{-v^{2}-u^{2}}{v^{2}} = \frac{2}{v^{2}}, \]

\[ y''_{xx} = \frac{2/v^{2}}{-e^{-t}v} = -\frac{2e^{t}}{\bigl(\cos t+\sin t\bigr)^{3}}. \tag{1} \]

Step 4: Differentiate (1) with respect to $t$

Product rule on $-2e^{t}v^{-3}$, with $v' = u$: $\bigl(y''_{xx}\bigr)'_{t} = -2e^{t}v^{-4}\bigl[v-3u\bigr]$.

The bracket is

\[ \bigl(\cos t+\sin t\bigr)-3\bigl(\cos t-\sin t\bigr) = -2\cos t+4\sin t = -2\bigl(\cos t-2\sin t\bigr), \]

\[ \bigl(y''_{xx}\bigr)'_{t} = \frac{4e^{t}\bigl(\cos t-2\sin t\bigr)}{\bigl(\cos t+\sin t\bigr)^{4}}. \tag{2} \]

Step 5: Divide by $\dot x$

\[ y'''_{xxx} = \frac{4e^{t}\bigl(\cos t-2\sin t\bigr)/\bigl(\cos t+\sin t\bigr)^{4}}{-e^{-t}\bigl(\cos t+\sin t\bigr)} = -\frac{4e^{2t}\bigl(\cos t-2\sin t\bigr)}{\bigl(\cos t+\sin t\bigr)^{5}}. \]

Step 6: Verification at $t = 0$

\[\dot x(0) = -1,\quad y'_{x}(0) = -1,\quad y''_{xx}(0) = -2 \quad\text{from (1)}\]

\[\text{From (2): } \bigl(y''_{xx}\bigr)'_{t}\Big|_{t=0} = \frac{4\cdot1\cdot1}{1} = 4, \qquad y'''_{xxx}(0) = \frac{4}{-1} = -4\]

\[\text{The boxed formula at } t = 0:\ -\frac{4\cdot1\cdot1}{1} = -4 \;\;\checkmark\]