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

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Answer 1 · 2017-06-19T14:58:12

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

Here the function is desccribed in parametric form as

#f(t)=(x(t),y(t))#, where #x(t)=t^2-1# and #y(t)=te^(2t^2-2)#

In such cases #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

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

and #(dy)/(dt)=1xxe^(2t^2-2)+txxe^(2t^2-2)xx4t#

= #e^(2t^2-2)(1+4t^2)#

Hence #(dy)/(dx)=(e^(2t^2-2)(1+4t^2))/(2t)#

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