What is the derivative of #f(t) = (t^2+1 , e^(-t^2-t) ) #?

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Answer 1 · 2017-03-07T13:39:54

#(dy)/(dx)=-(2t+1)/(2t)e^(-t^2-t)#

When a function is given in parametric form such as #f(t)=(x(t),y(t))#,

derivative #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

As #f(t)=(t^2+1,e^(-t^2-t))#

#(dy)/(dt)=e^(-t^2-t)xx(-2t-1)#

= #-(2t+1)e^(-t^2-t)#

and #(dx)/(dt)=2t#

Hence #(dy)/(dx)=-(2t+1)/(2t)e^(-t^2-t)#