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

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Answer 1 · 2016-01-14T02:55:50

Integrate each part seperately, since they are in a different axis each.

#f'(t)=(2t-cost,-1/(t-1)^2)#

1st part

#(t^2-sint)'=2t-cost#

2nd part

#(1/(t-1))'=((t-1)^-1)'=-1*(t-1)^(-1-1)*(t-1)'=#

#=-(t-1)^(-2)*1=-1/(t-1)^2#

Result

#f'(t)=(2t-cost,-1/(t-1)^2)#

Answer 2 · 2016-01-14T02:57:28

#-1/((2t-cost)(t-1)^2)#

#x(t)=t^2-sint#
#y(t)=1/(t-1)#

#x'(t)=2t-cost#
#y'(t)=-1/(t-1)^2#

To find the derivative of a parametric function, find

#dy/dx=(dy/dt)/(dx/dt)=(y'(t))/(x'(t))=(-1/(t-1)^2)/(2t-cost)=-1/((2t-cost)(t-1)^2)#