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How do you differentiate x+siny=0 with respect to x?

Thank you so much!!

Answer 1

Isaiah Allen

#(d y)/(d x) = -1/(sqrt (1-x^2))#

#x + sin y = 0# #sin y = -x# #y = sin^(-1) (-x)#

let say # u = -x#, #(d u)/(d x) = -1#

therefore,

#y = sin^(-1) u# #(d y)/(d x) = 1/(sqrt (1-u^2)) * (du)/(dx)#

#(d y)/(d x) = 1/(sqrt (1-(-x)^2)) * (-1)#

#(d y)/(d x) = -1/(sqrt (1-x^2))#

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Answer 2

Luke Alvarez

To differentiate ( x + \sin(y) = 0 ) with respect to ( x ), you can use implicit differentiation.

First, differentiate each term with respect to ( x ):

( \frac{d}{dx}[x] = 1 )

( \frac{d}{dx}[\sin(y)] = \cos(y) \frac{dy}{dx} ) (by the chain rule)

Combine the derivatives and set the result equal to zero:

( 1 + \cos(y) \frac{dy}{dx} = 0 )

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

( \frac{dy}{dx} = -\frac{1}{\cos(y)} )

Thus, the derivative of ( x + \sin(y) = 0 ) with respect to ( x ) is ( \frac{dy}{dx} = -\frac{1}{\cos(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.