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

Answer 1

#= - 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#
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Answer 2

To differentiate the given parametric equations ( x(t) = \frac{1}{t} ) and ( y(t) = \ln(t) ) with respect to ( t ), you would differentiate each equation separately using the chain rule and the rules of differentiation for natural logarithm.

( \frac{dx}{dt} = \frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2} )

( \frac{dy}{dt} = \frac{d}{dt}\left(\ln(t)\right) = \frac{1}{t} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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