What is (\partial)/(\partial x)(x^4y^3-4x^2y+y^5) ?

The answer to (\partial)/(\partial x)(x^4y^3-4x^2y+y^5) is 4x^3y^3-8xy

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(\partial)/(\partial x)(x^4y^3-4x^2y+y^5)

\frac{\partial}{\partial x} (x^{4} y^{3} - 4 x^{2} y + y^{5})

4 x^{3} y^{3} - 8 x y

Solution steps

\frac{\partial}{\partial x} (x^{4} y^{3} - 4 x^{2} y + y^{5})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4} y^{3}) - \frac{\partial}{\partial x} (4 x^{2} y) + \frac{\partial}{\partial x} (y^{5})

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

\frac{\partial}{\partial x} (4 x^{2} y) = 8 y x

\frac{\partial}{\partial x} (y^{5}) = 0

= 4 y^{3} x^{3} - 8 y x + 0

Simplify

= 4 x^{3} y^{3} - 8 x y

f (x) = cos (\frac{π x}{2}), at x = 0.8

f (x) = 2 x^{4} e^{x}

\int \frac{x ^{3}}{3} - \frac{5 x ^{2}}{2} + 6 x + C d x

\int \frac{1}{200 + 3 t} d t

n = 3 \sum \infty \frac{4}{n ^{2} - 1}

What is (\partial)/(\partial x)(x^4y^3-4x^2y+y^5) ?
The answer to (\partial)/(\partial x)(x^4y^3-4x^2y+y^5) is 4x^3y^3-8xy