# How do you find the derivative of #y=ln((x-1)/(x+1))^(1/3)#?

# dy/dx = 2/(3x^2-3)#

Differentiating the LHS implicity and the RHS using the quotient rule gives:

Hence,

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So

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To find the derivative of y = ln((x - 1)/(x + 1))^(1/3), first use the chain rule. The derivative of ln(u) is (1/u) * u', where u = ((x - 1)/(x + 1))^(1/3). Then, find the derivative of u using the chain rule and simplify. The derivative of u is [(1/3) * ((x - 1)/(x + 1))^(-2/3) * ((1/(x + 1)) - (1/(x - 1)))]. Finally, multiply the derivatives of ln(u) and u together to get the derivative of y.

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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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