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How do you differentiate the following parametric equation: # x(t)=e^t/(t+t), y(t)=t-e^(3t) #?

Answer 1

Christopher Agresta

# dy/dx = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1) #

I assume that you want more than just the derivatives of the parametric equation, and actually require #dy/dx#

We have;

# x(t) = e^t/(t+t) = e^t/(2t)#

Differentiating wrt #t#, and applying the quotient rule, we have;

# dx/dt = {(2t)(e^t) - (2)(e^t)}/(2t)^2 # # " " = (2(t-1)e^t)/(4t^2) # # " " = ((t-1)e^t)/(2t^2) #

And:

# y(t)=t-e^(3t) #

Differentiating wrt #t#, and applying the chain rule, we have;

# dy/dt = 1 - 3e^(3t) #

And finally by the chain rule:

# dy/dx = (dy//dt) / (dx//dt) # # " " = (1 - 3e^(3t)) / (((t-1)e^t)/(2t^2)) # # " " = (1 - 3e^(3t)) * (2t^2)/((t-1)e^t) # # " " = (2t^2(1 - 3e^(3t))e^(-t)) /((t-1) #

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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.