How do you differentiate the following parametric equation: # x(t)=t/(t-4), y(t)=1/(1-t^2) #?

1 Answer
Feb 28, 2018

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

Explanation:

#dy/dx=(y'(t))/(x'(t))#

#y(t)=1/(1-t^2)#
#y'(t)=((1-t^2)d/dt[1]-1d/dt[1-t^2])/(1-t^2)^2#
#color(white)(y'(t))=(-(-2t))/(1-t^2)^2#
#color(white)(y'(t))=(2t)/(1-t^2)^2#

#x(t)=t/(t-4)#
#x'(t)=((t-4)d/dt[t]-t d/dt[t-4])/(t-4)^2#
#color(white)(x'(t))=(t-4-t)/(t-4)^2#
#color(white)(x'(t))=-4/(t-4)^2#

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