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

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Answer 1 · 2016-08-08T11:19:12

#= - t =- 1/x#

assuming youre looking for #dy/dx#, you use the following fact

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

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

#= - t #

and if you like

#=- 1/x# as #x = 1/t#

we can check this by de-parameterising the equation

from #t = 1/x# we have #y = ln (1/x) = - ln x# so #dy/dx = - 1/x#