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How do you differentiate #sinx-siny=x-y-5#?

Answer 1

# dy/dx = (cosx - 1)/(cosy - 1)#

# sinx-siny=x-y-5 #

Differentiate wrt #x# # :. d/dxsinx-d/dxsiny=d/dxx - d/dxy - d/dx5 # # :. cosx - d/dxsiny=1 - dy/dx #

We can't differentiate #siny# wrt #x# but we can differentiate wet y using the chain rule (this is the implicit differentiation)

# :. cosx - dy/dxd/dysiny=1 - dy/dx #

# :. cosx - dy/dxcosy=1 - dy/dx #

We can now collect terms and factorise: # :. dy/dxcosy - dy/dx = cosx - 1#

# :. dy/dx(cosy - 1) = cosx - 1#

# :. dy/dx = (cosx - 1)/(cosy - 1)#

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Answer 2

Adam Adkins

To differentiate the equation ( \sin(x) - \sin(y) = x - y - 5 ), you can use implicit differentiation. The derivative of ( \sin(x) ) is ( \cos(x) ) and the derivative of ( \sin(y) ) is ( \cos(y) ). Therefore, the derivative of the left side with respect to ( x ) is ( \cos(x) ) and the derivative of the right side with respect to ( x ) is ( 1 ). Similarly, the derivative of the left side with respect to ( y ) is ( -\cos(y) ) and the derivative of the right side with respect to ( y ) is ( -1 ). So, the differentiated equation is ( \cos(x) - \cos(y) = 1 ).

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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.