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

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Answer 1 · 2016-08-03T14:05:27

Hence derivative of #f(t)=(t/(t+1),te^(2t-1))# is #(t+1)^2(1+2t)e^(2t-1)#

In parametric form of function where #f(t)=(g(t),h(t))#

#(dy)/(dx)=((dh)/(dt))/((dg)/(dt))#

as such derivative of given function #f(t)=(t/(t+1),te^(2t-1))# is

#(d/dt(te^(2t-1)))/(d/(dt)(t/(t+1))#

#d/dt(te^(2t-1))=1xxe^(2t-1)+2txxe^(2t-1)=e^(2t-1)(1+2t)#

and #d/(dt)(t/(t+1))=d/(dt)(1-1/(t+1))=1/(t+1)^2#

Hence derivative of #f(t)=(t/(t+1),te^(2t-1))# is

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