# How do you find the derivative of #y = secx/(1 + tanx)#?

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To find the derivative of ( y = \frac{\sec x}{1 + \tan x} ), we can use the quotient rule.

The quotient rule states that for functions ( u(x) ) and ( v(x) ) where ( v(x) \neq 0 ), the derivative of ( \frac{u(x)}{v(x)} ) is given by:

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

Using the quotient rule, the derivative of ( y ) with respect to ( x ) is:

[ y' = \frac{(\sec x)'(1 + \tan x) - \sec x(\tan x)'}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x \tan x (1 + \tan x) - \sec x \sec^2 x}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x \tan x + \sec^2 x \tan x - \sec x \sec^2 x}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x \tan x + \sec^3 x \tan x - \sec^3 x}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x (\tan x + \sec^2 x \tan x - \sec^2 x)}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x (\tan x(1 + \sec^2 x) - \sec^2 x)}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x (\tan x + \sec^2 x - \sec^2 x)}{(1 + \tan x)^2} ]

[ y' = \frac{\sec x \tan x}{(1 + \tan x)^2} ]

So, the derivative of ( y ) with respect to ( x ) is ( \frac{\sec x \tan x}{(1 + \tan 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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