What is (\partial)/(\partial y)(8xy-1/4 (x+y)^4) ?

The answer to (\partial)/(\partial y)(8xy-1/4 (x+y)^4) is 8x-(x+y)^3

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(\partial)/(\partial y)(8xy-1/4 (x+y)^4)

\frac{\partial}{\partial y} (8 x y - \frac{1}{4} (x + y)^{4})

8 x - (x + y)^{3}

Solution steps

\frac{\partial}{\partial y} (8 x y - \frac{1}{4} (x + y)^{4})

Simplify

8 x y - \frac{1}{4} (x + y)^{4} : 8 x y - \frac{( x + y ) ^{4}}{4}

= \frac{\partial}{\partial y} (8 x y - \frac{( x + y ) ^{4}}{4})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (8 x y) - \frac{\partial}{\partial y} (\frac{( x + y ) ^{4}}{4})

\frac{\partial}{\partial y} (8 x y) = 8 x

\frac{\partial}{\partial y} (\frac{( x + y ) ^{4}}{4}) = (x + y)^{3}

= 8 x - (x + y)^{3}

x \to x lim (sin (x))

y = x^{2}, y = 9 x, y = - 1

\frac{d}{d x} (- x e^{- x} - e^{- x} + 1)

\int \frac{4 x ^{3} - 2 x + 1}{x} d x

y = 3 x^{3} - x^{2} + 1, (2, 21)

What is (\partial)/(\partial y)(8xy-1/4 (x+y)^4) ?
The answer to (\partial)/(\partial y)(8xy-1/4 (x+y)^4) is 8x-(x+y)^3