Question:

$\color{blue}{\text{Step 1:}}$ We are given $x=t^5-5t^3-20t+7$ and $y=4t^3-3t^2-18t+3$. To find $\frac{dy}{dx}$, we will use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $$ \begin{aligned} \frac{dy}{dt} &= 12t^2 - 6t - 18 \\ \frac{dx}{dt} &= 5t^4 - 15t^2 - 20 \end{aligned} $$ $\color{blue}{\text{Step 2:}}$ Now, we can find $\frac{dy}{dx}$ using the formula. $$ \frac{dy}{dx} = \frac{12t^2 - 6t - 18}{5t^4 - 15t^2 - 20} $$ $\color{blue}{\text{Step 3:}}$ We need to find $\left[\frac{dy}{dx}\right]_{t=1}$. $$ \begin{aligned} \left[\frac{dy}{dx}\right]_{t=1} &= \frac{12(1)^2 - 6(1) - 18}{5(1)^4 - 15(1)^2 - 20} \\ &= \frac{12 - 6 - 18}{5 - 15 - 20} \\ &= \frac{-12}{-30} \\ &= \frac{2}{5} \end{aligned} $$ $\color{blue}{\text{Step 4:}}$ Finally, we need to find $-5 \times \frac{dy}{dx}$ at $t=1$. $$ \begin{aligned} \left[-5 \frac{dy}{dx}\right]_{t=1} &= -5 \times \frac{2}{5} \\ &= -2 \end{aligned} $$ Therefore: $-2$