# How do you use the chain rule to differentiate #y=(1/(t-3))^2#?

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To differentiate ( y = \left(\frac{1}{t-3}\right)^2 ) using the chain rule, follow these steps:

- Let ( u = \frac{1}{t-3} ).
- Differentiate ( u ) with respect to ( t ) to get ( \frac{du}{dt} ).
- Apply the chain rule: ( \frac{dy}{dt} = 2u \cdot \frac{du}{dt} ).
- Substitute the expression for ( u ) and ( \frac{du}{dt} ) into the equation.
- Simplify the result to get the final answer for ( \frac{dy}{dt} ).

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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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