# How do you differentiate the following parametric equation: # x(t)=3(t+1)^2+2e^t, y(t)= (t+2)^2+t^2#?

#(dy)/(dx) = (3(t+1))/(2t+1)#

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To differentiate the given parametric equations (x(t)) and (y(t)), we will use the chain rule and the product rule where necessary.

[ x(t) = 3(t+1)^2 + 2e^t ] [ y(t) = (t+2)^2 + t^2 ]

[ \frac{dx}{dt} = 3 \cdot 2(t+1) \cdot \frac{d(t+1)}{dt} + 2 \cdot \frac{d(e^t)}{dt} ] [ \frac{dy}{dt} = 2(t+2) \cdot \frac{d(t+2)}{dt} + 2t ]

[ \frac{dx}{dt} = 6(t+1) + 2e^t ] [ \frac{dy}{dt} = 2(t+2) + 2t ]

So, the derivatives with respect to (t) are:

[ \frac{dx}{dt} = 6(t+1) + 2e^t ] [ \frac{dy}{dt} = 2(t+2) + 2t ]

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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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