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

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

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

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

20 y^{4} - 2 x y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{2}) = 0

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

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

= 0 - 2 x y + 20 y^{4}

Simplify

= 20 y^{4} - 2 x y

\frac{\partial}{\partial y} (sin (x) cos (y))

\frac{d}{d x} (\frac{3}{x ^{2} + 2})

\frac{d}{d x} (x - 2 y + 2 z)

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

\frac{1}{( s - 1 ) ^{2} + 1}

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