What is (\partial)/(\partial x)((x+y)^p-bx-(1-b)y) ?

The answer to (\partial)/(\partial x)((x+y)^p-bx-(1-b)y) is -b+p(x+y)^{p-1}

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(\partial)/(\partial x)((x+y)^p-bx-(1-b)y)

\frac{\partial}{\partial x} ((x + y)^{p} - b x - (1 - b) y)

- b + p (x + y)^{p - 1}

Solution steps

\frac{\partial}{\partial x} ((x + y)^{p} - b x - (1 - b) y)

Treat

p, y, b

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial x} ((x + y)^{p}) = p (x + y)^{p - 1}

\frac{\partial}{\partial x} (b x) = b

\frac{\partial}{\partial x} ((1 - b) y) = 0

= p (x + y)^{p - 1} - b - 0

Simplify

= - b + p (x + y)^{p - 1}

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What is (\partial)/(\partial x)((x+y)^p-bx-(1-b)y) ?
The answer to (\partial)/(\partial x)((x+y)^p-bx-(1-b)y) is -b+p(x+y)^{p-1}