What is (\partial)/(\partial x)(3xy^2+20x^2y^3) ?

The answer to (\partial)/(\partial x)(3xy^2+20x^2y^3) is 3y^2+40y^3x

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(\partial)/(\partial x)(3xy^2+20x^2y^3)

\frac{\partial}{\partial x} (3 x y^{2} + 20 x^{2} y^{3})

3 y^{2} + 40 y^{3} x

Solution steps

\frac{\partial}{\partial x} (3 x y^{2} + 20 x^{2} y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x y^{2}) + \frac{\partial}{\partial x} (20 x^{2} y^{3})

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

\frac{\partial}{\partial x} (20 x^{2} y^{3}) = 40 y^{3} x

= 3 y^{2} + 40 y^{3} x

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{4}) = 2, y^{^{'}} (\frac{π}{4}) = - 2

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

y^{^{''}} = 1 + (y^{^{'}})^{2}

\frac{d ^{4}}{d x ^{4}} (11 e^{x^{2}})

What is (\partial)/(\partial x)(3xy^2+20x^2y^3) ?
The answer to (\partial)/(\partial x)(3xy^2+20x^2y^3) is 3y^2+40y^3x