# Find the maxima for #x/(1+xtanx)#?

We have a local maxima at

graph{(y-x)(y-cosx)=0 [-0.667, 1.833, -0.13, 1.12]}

graph{x/(1+xtanx) [-0.0256, 1.2155, -0.0604, 0.5603]}

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To find the maxima of the function ( \frac{x}{1 + x\tan(x)} ), we first need to find its derivative and then solve for where the derivative equals zero.

The derivative of the function with respect to ( x ) can be found using the quotient rule:

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

Simplify the expression:

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

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

Set the derivative equal to zero and solve for ( x ):

[ \frac{1 - x\tan^2(x)}{(1 + x\tan(x))^2} = 0 ]

Since the numerator is the only part that can equal zero, we have:

[ 1 - x\tan^2(x) = 0 ]

[ 1 = x\tan^2(x) ]

[ x = \frac{1}{\tan^2(x)} ]

[ x = \cot^2(x) ]

Therefore, the maxima occur at the solutions to the equation ( x = \cot^2(x) ). These solutions can be found numerically or graphically.

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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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