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

Answer 1

Christopher Agresta

#[dy/dx]_(2,pi/3) = 1/(2sqrt3)#

Differentiate the equation with respect to #x#:

#d/dx (xcosy) = 0#

#cosy -xsinydy/dx = 0#

#cosy = xsinydy/dx #

#dy/dx = 1/(xtgy)#

In the point #(2,pi/3)# the value is then:

#[dy/dx]_(2,pi/3) = 1/(2tg(pi/3)) = 1/(2sqrt3)#

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Answer 2

Scarlett Allison

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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Answer from HIX Tutor

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.