What is (\partial)/(\partial b)((ax+b-y)^2) ?

The answer to (\partial)/(\partial b)((ax+b-y)^2) is 2(ax+b-y)

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(\partial)/(\partial b)((ax+b-y)^2)

\frac{\partial}{\partial b} ((a x + b - y)^{2})

2 (a x + b - y)

Solution steps

\frac{\partial}{\partial b} ((a x + b - y)^{2})

Treat

a, x, y

as constants

Apply the chain rule:

2 (a x + b - y) \frac{\partial}{\partial b} (a x + b - y)

= 2 (a x + b - y) \frac{\partial}{\partial b} (a x + b - y)

\frac{\partial}{\partial b} (a x + b - y) = 1

= 2 (a x + b - y) \cdot 1

Simplify

= 2 (a x + b - y)

\int e^{\frac{- x}{3}} cos (5 x) d x

x \to 0 lim (\frac{11 sin ( 8 x )}{x})

7 y^{3} + 9 x^{8} = 5 y + 11 x, (1, 1)

x^{4} + 5 x^{2} - x

f (x) = 8 e^{e}

What is (\partial)/(\partial b)((ax+b-y)^2) ?
The answer to (\partial)/(\partial b)((ax+b-y)^2) is 2(ax+b-y)