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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{d}{d x} (cos (\frac{π}{x}))

\int_{2}^{\infty} - 8 x^{- 2} d x

x \to 0 lim (\frac{cos ( 3 x )}{x})

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

\frac{d y}{d x} = \frac{y ^{2} + 1}{x ^{2} + 1}

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