# How do you find #(d^2y)/(dx^2)# given #y^2=(x-1)/(x+1)#?

# (d^2y)/dx^2 = -(2x-1) / ((x+1)^3(x-1)) * 1/sqrt((x-1)/(x+1))#

Alternatively:

# (d^2y)/dx^2 = -(2x-1) / ((x+1)^4((x-1)/(x+1))^(3/2)) #

We have established:

Differentiating the LHS of [1] Implicitly with the product rule, and the RHS using the chain rule we get:

We can also write this as;

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To find (\frac{{d^2y}}{{dx^2}}), first, express (y) explicitly in terms of (x), then differentiate twice with respect to (x), using the chain rule and product rule where necessary.

Given (y^2 = \frac{{x - 1}}{{x + 1}}), solve for (y) to get (y = \pm \sqrt{\frac{{x - 1}}{{x + 1}}}).

Taking the derivative of (y) with respect to (x) gives:

(\frac{{dy}}{{dx}} = \pm \frac{1}{2} \left(\frac{{x - 1}}{{x + 1}}\right)^{-\frac{1}{2}} \cdot \frac{{(x + 1) - (x - 1)}}{{(x + 1)^2}})

Now, differentiate (\frac{{dy}}{{dx}}) with respect to (x) to find (\frac{{d^2y}}{{dx^2}}).

This process involves using the quotient rule and the chain rule. After simplification, you should arrive at the final expression for (\frac{{d^2y}}{{dx^2}}).

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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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