find the derivative of y = te^(-t^5) dy/dt =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $t$, then $\frac{dy}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Here, $u = t$ and $v = e^{-t^{5}}$.

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

Since $u = t$, then $\frac{du}{dt}=1$.

Step3: Find $\frac{dv}{dt}$

Let $u=-t^{5}$, then $v = e^{u}$. By the chain - rule $\frac{dv}{dt}=\frac{dv}{du}\cdot\frac{du}{dt}$. We know that $\frac{dv}{du}=e^{u}$ and $\frac{du}{dt}=-5t^{4}$. So $\frac{dv}{dt}=e^{-t^{5}}\cdot(- 5t^{4})=-5t^{4}e^{-t^{5}}$.

Step4: Substitute into product - rule

$\frac{dy}{dt}=t\cdot(-5t^{4}e^{-t^{5}})+e^{-t^{5}}\cdot1$. Simplify the expression: $\frac{dy}{dt}=e^{-t^{5}}(1 - 5t^{5})$.

Answer:

$e^{-t^{5}}(1 - 5t^{5})$