# How do you derive #y = (sinx)/[1 - sin^2(x))# using the quotient rule?

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To derive ( y = \frac{\sin(x)}{1 - \sin^2(x)} ) using the quotient rule, we first identify ( u(x) = \sin(x) ) and ( v(x) = 1 - \sin^2(x) ). Then, we use the quotient rule formula:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} ]

Now, we find the derivatives of ( u(x) ) and ( v(x) ):

[ u'(x) = \cos(x) ] [ v'(x) = -2\sin(x)\cos(x) ]

Substituting these into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{\sin(x)}{1 - \sin^2(x)} \right) = \frac{(1 - \sin^2(x)) \cdot \cos(x) - \sin(x) \cdot (-2\sin(x)\cos(x))}{(1 - \sin^2(x))^2} ]

[ = \frac{\cos(x) - \sin^2(x)\cos(x) + 2\sin^2(x)\cos(x)}{(1 - \sin^2(x))^2} ]

[ = \frac{\cos(x) + \sin^2(x)\cos(x)}{(1 - \sin^2(x))^2} ]

[ = \frac{\cos(x) + \sin(x)\cos^2(x)}{(1 - \sin^2(x))^2} ]

So, the derivative of ( y = \frac{\sin(x)}{1 - \sin^2(x)} ) with respect to ( x ) using the quotient rule is:

[ \frac{d}{dx} \left( \frac{\sin(x)}{1 - \sin^2(x)} \right) = \frac{\cos(x) + \sin(x)\cos^2(x)}{(1 - \sin^2(x))^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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