What is the derivative of # f(x)=cosx/(1+sin^2x)#?
Examining each section separately
whereby you obtain
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To find the derivative of ( f(x) = \frac{\cos(x)}{1 + \sin^2(x)} ), use the quotient rule.
Let ( u(x) = \cos(x) ) and ( v(x) = 1 + \sin^2(x) ). Then, applying the quotient rule:
[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]
[ f'(x) = \frac{-\sin(x)(1 + \sin^2(x)) - \cos(x)(2\sin(x)\cos(x))}{(1 + \sin^2(x))^2} ]
[ f'(x) = \frac{-\sin(x) - \sin^3(x) - 2\sin(x)\cos^2(x)}{(1 + \sin^2(x))^2} ]
[ f'(x) = \frac{-\sin(x) - \sin^3(x) - 2\sin(x)(1 - \sin^2(x))}{(1 + \sin^2(x))^2} ]
[ f'(x) = \frac{-\sin(x) - \sin^3(x) - 2\sin(x) + 2\sin^3(x)}{(1 + \sin^2(x))^2} ]
[ f'(x) = \frac{-3\sin(x) + \sin^3(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.
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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