How do I simplify this calculus problem?

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SUMMARY

The discussion focuses on simplifying the calculus equation 2x + 3 (y-x)^2 * (y'-1) = 0 to the form 2x + 3 (y-x)^2 * y' - 3 (y-x)^2 = 0. The key transformation involves applying the distributive property, specifically 3a(b - c) = 3ab - 3ac, where a = (y - x)^2, b = y', and c = 1. This simplification is crucial for solving the equation effectively.

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tmt1
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Hi I'm working on a problem for calculus

and I got this

2x + 3 (y-x)^2 * (y'-1) = 0 .

And then, in the answer part it says I have to get to here, which I'm not sure how to do.

2x + 3 (y-x)^2 * y'- 3 (y-x)^2 = 0 ,

Thanks,

Tim
 
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tmt said:
Hi I'm working on a problem for calculus

and I got this

2x + 3 (y-x)^2 * (y'-1) = 0 .

And then, in the answer part it says I have to get to here, which I'm not sure how to do.

2x + 3 (y-x)^2 * y'- 3 (y-x)^2 = 0 ,

Thanks,

Tim

It's \displaystyle \begin{align*} 3a\left( b - c \right) = 3a\,b - 3a\,c \end{align*}, where \displaystyle \begin{align*} a = \left( y - x \right) ^2 , b = y', c = 1 \end{align*}.
 

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