Question

find ( dy/dx ) in terms of ( t ) if ( x = te^{t}, y = -2t - 2e^{t} )

Explanation:

Step1: Find $\frac{dy}{dt}$

Given $y=-2t - 2e^{t}$, we differentiate $y$ with respect to $t$ using the sum rule and the derivative of exponential function. The derivative of $-2t$ with respect to $t$ is $-2$, and the derivative of $-2e^{t}$ with respect to $t$ is $-2e^{t}$. So, $\frac{dy}{dt}=-2 - 2e^{t}$.

Step2: Find $\frac{dx}{dt}$

Given $x = te^{t}$, we use the product rule for differentiation. The product rule states that if $u$ and $v$ are functions of $t$, then the derivative of $uv$ is $u\frac{dv}{dt}+v\frac{du}{dt}$. Let $u = t$ and $v = e^{t}$. Then $\frac{du}{dt}=1$ and $\frac{dv}{dt}=e^{t}$. So, $\frac{dx}{dt}=t\times e^{t}+e^{t}\times1=e^{t}(t + 1)$.

Step3: Find $\frac{dy}{dx}$

We know that $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Substituting the values of $\frac{dy}{dt}$ and $\frac{dx}{dt}$ we found above, we get $\frac{dy}{dx}=\frac{-2 - 2e^{t}}{e^{t}(t + 1)}$.

Answer:

$\frac{-2 - 2e^{t}}{e^{t}(t + 1)}$