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

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Answer 1 · 2017-07-21T07:11:03

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

${(y - 1)}^{2} + x = 0 \Rightarrow \frac{d {(y - 1)}^{2}}{d x} + \frac{d x}{d x} = 0 \Rightarrow$ $(y-1)^2+x=0=> (d(y-1)^2)/dx+dx/dx=0=>$

$2 (y - 1) \frac{d y}{d x} + 1 = 0 \Rightarrow \frac{d y}{d x} = - \frac{1}{2 (y - 1)}$ $2(y-1)dy/dx+1=0 =>dy/dx=-1/(2(y-1)$

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