What is (\partial)/(\partial x)(5x^4+x^3y+xy+y^4) ?

The answer to (\partial)/(\partial x)(5x^4+x^3y+xy+y^4) is 20x^3+3x^2y+y

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(\partial)/(\partial x)(5x^4+x^3y+xy+y^4)

\frac{\partial}{\partial x} (5 x^{4} + x^{3} y + x y + y^{4})

20 x^{3} + 3 x^{2} y + y

Solution steps

\frac{\partial}{\partial x} (5 x^{4} + x^{3} y + x y + y^{4})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (5 x^{4}) + \frac{\partial}{\partial x} (x^{3} y) + \frac{\partial}{\partial x} (x y) + \frac{\partial}{\partial x} (y^{4})

\frac{\partial}{\partial x} (5 x^{4}) = 20 x^{3}

\frac{\partial}{\partial x} (x^{3} y) = 3 y x^{2}

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

\frac{\partial}{\partial x} (y^{4}) = 0

= 20 x^{3} + 3 y x^{2} + y + 0

Simplify

= 20 x^{3} + 3 x^{2} y + y

\frac{\partial}{\partial y} (4 x y^{2})

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

\frac{d}{d x} (4 sec (7 x))

\frac{d}{d x} (22 - x)

\int \frac{1}{y + y ^{2}} d y

What is (\partial)/(\partial x)(5x^4+x^3y+xy+y^4) ?
The answer to (\partial)/(\partial x)(5x^4+x^3y+xy+y^4) is 20x^3+3x^2y+y