# How do you find #dy/dx# by implicit differentiation of #xcosy=1# and evaluate at point (2, pi/3)?

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To find ( \frac{{dy}}{{dx}} ) by implicit differentiation of ( x \cos(y) = 1 ), differentiate both sides of the equation with respect to ( x ), then solve for ( \frac{{dy}}{{dx}} ).

( x \cos(y) = 1 )

Differentiating both sides with respect to ( x ):

( \frac{{d}}{{dx}}(x \cos(y)) = \frac{{d}}{{dx}}(1) )

Using the product rule on the left side:

( \cos(y) - x \sin(y) \frac{{dy}}{{dx}} = 0 )

Now, solve for ( \frac{{dy}}{{dx}} ):

( \frac{{dy}}{{dx}} = \frac{{\sin(y)}}{{\cos(y)}} = \tan(y) )

To evaluate at the point ( (2, \frac{\pi}{3}) ), substitute ( x = 2 ) and ( y = \frac{\pi}{3} ) into the derivative expression:

( \frac{{dy}}{{dx}} = \tan(\frac{\pi}{3}) = \sqrt{3} )

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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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