What is the derivative of #f(t) = (t/(t+1) , t/(4t^2-t) ) #?

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Answer 1 · 2016-09-28T06:49:09

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

#f(t)=(t/(t+1),t/(4t^2-t))# is parametric form of equation,

where #t/(t+1)# represents #x# and #t/(4t^2-t)# represents #y#.

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

Here we can find #(dy)/(dt)# and #(dx)/(dt)# using quotient rule

#(dy)/(dt)=(1xx(4t^2-t)-txx(8t-1))/(4t^2-t)^2=(4t^2-t-8t^2+t)/(4t^2-t)^2=(-4t^2)/(4t^2-t)^2#

#(dx)/(dt)=(1xx(t+1)-txx1)/(t+1)^2=(t+1-t)/(t+1)^2=1/(t+1)^2#

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

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