Question 381641
{{{df/dt=(df/dx)*(dx/dt)+(df/dy)*(dy/dt)}}}
{{{df/dt=(2e^(2t^2-3/t))*(2t)+(-3)(e^(2t^2-3/t))(-t^(-2))}}}
{{{df/dt=4t*e^(2t^2-3/t)+3/t^2e^(2t^2-3/t)}}}
{{{df/dt=e^(2t^2-3/t)(4t+3/t^2)}}}
When {{{t=1}}}
{{{df/dt=e^(2-3)(4+3)}}}
{{{highlight(df/dt=7/e)}}}
I get that too.
Well done.