Question

What is the derivative of `f(t) = (e^t/t +e^t, e^t-tcost )`?

Answer 1

`f'(t)=(e^t((t-1)/t^2+1),e^t+tsint-cost)`

Explanation:

Let `x(t)=e^t/t+e^t` and `y(t)=e^t-tcost` Therefore, `f(t)=(x(t),y(t))`, so `f'(t)=(x'(t),y'(t))`.

`x'(t)=d/dt[e^t/t+e^t]`
`x'(t)=d/dt[e^t/t]+d/dt[e^t]`
`x'(t)=(td/dt[e^t]-e^td/dt[t])/t^2+e^t`
`x'(t)=(te^t-e^t)/t^2+e^t`
`x'(t)=e^t((t-1)/t^2+1)`

`y'(t)=d/dt[e^t-tcost]`
`y'(t)=d/dt[e^t]-d/dt[tcost]`
`y'(t)=e^t-(td/dt[cost]+costd/dt[t])`
`y'(t)=e^t-(-tsint+cost)`
`y'(t)=e^t+tsint-cost`

`f'(t)=(e^t((t-1)/t^2+1),e^t+tsint-cost)`