# If # x = (4t^2)/(5t^2+6) #, and # y = t^3 # then find # dy/dx#?

# dy/dx = (t(5t^2+6)^2) / (16) #

We have:

Then, By the chain rule, we have:

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To find ( \frac{dy}{dx} ), differentiate ( y ) with respect to ( t ) and ( x ) with respect to ( t ), then use the chain rule to find ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} ]

Given ( y = t^3 ), ( \frac{dy}{dt} = 3t^2 ).

Given ( x = \frac{4t^2}{5t^2 + 6} ), ( \frac{dx}{dt} = \frac{d}{dt} \left( \frac{4t^2}{5t^2 + 6} \right) ).

Use the quotient rule to differentiate:

[ \frac{d}{dt} \left( \frac{4t^2}{5t^2 + 6} \right) = \frac{(8t)(5t^2 + 6) - (4t^2)(10t)}{(5t^2 + 6)^2} ]

Now substitute both derivatives into ( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} ):

[ \frac{dy}{dx} = \frac{3t^2}{\frac{(8t)(5t^2 + 6) - (4t^2)(10t)}{(5t^2 + 6)^2}} ]

Simplify this expression to get the final answer for ( \frac{dy}{dx} ).

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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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