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

Let's first change the original expression to make it easier to differentiate:

We start by differentiating the expression outside the brackets to get:

We then differentiate the what is inside the brackets:

Now by the chain rule we just multiply these two expressions together to get:

This is the final answer.

So that the final step by the chain rule is:

This is the final answer.

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To differentiate ( y = \frac{1}{t^2 + 3x - 1} ) using the chain rule, you first rewrite the function as ( y = (t^2 + 3x - 1)^{-1} ). Then, you differentiate with respect to ( t ) treating ( t^2 + 3x - 1 ) as the inner function and ( u = t^2 + 3x - 1 ) as the outer function. The derivative of ( u ) with respect to ( t ) is ( \frac{du}{dt} = 2t ). Then, you apply the chain rule, which states that ( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} ). So, ( \frac{dy}{du} = -1 \cdot (t^2 + 3x - 1)^{-2} ) and ( \frac{dy}{dt} = -\frac{2t}{(t^2 + 3x - 1)^2} ).

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