How do you differentiate #(y-1)^2+x=0#?

Answer 1

Christopher Allers

dy/dx=-1/(2(y-1)#

#(y-1)^2+x=0=> (d(y-1)^2)/dx+dx/dx=0=>#

#2(y-1)dy/dx+1=0 =>dy/dx=-1/(2(y-1)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

William Abdalla

To differentiate ( (y - 1)^2 + x = 0 ) implicitly with respect to ( x ), follow these steps:

Differentiate each term of the equation with respect to ( x ).
For the term ( (y - 1)^2 ), you need to use the chain rule. The derivative of ( (y - 1)^2 ) with respect to ( x ) is ( 2(y - 1) \frac{dy}{dx} ).
For the term ( x ), the derivative is ( 1 ).
Set up the equation with the derivatives and solve for ( \frac{dy}{dx} ).

[ 2(y - 1) \frac{dy}{dx} + 1 = 0 ]

[ 2(y - 1) \frac{dy}{dx} = -1 ]

[ \frac{dy}{dx} = \frac{-1}{2(y - 1)} ]

By signing up, you agree to our Terms of Service and Privacy Policy

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.