What is the derivative of #cos(x-y)=0#?

Answer 1

Charles Anctil

#dy/(dx)=1#

Given #f(x,y)=0#

#df=f_xdx+f_ydy=0# so

#dy/(dx)=-f_x/(f_y)= sin(x-y)/sin(x-y)=1#

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Answer 2

Ezekiel Alexander

#dy/dx=1#

We can rearrange from the beginning:

#x-y=arccos(0)#

#x-y=pi/2#

#y=x-pi/2#

#dy/dx=1#

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Answer 3

Isaiah Allen

To find the derivative of ( \cos(x-y) = 0 ), you can use implicit differentiation. The derivative with respect to ( x ) of both sides yields:

[ -\sin(x-y) \cdot (1 - \frac{dy}{dx}) = 0 ]

Solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \sin(x-y) ]

So, the derivative of ( \cos(x-y) = 0 ) with respect to ( x ) is ( \sin(x-y) ).

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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.