What is (\partial)/(\partial x)(y^2ln(y)-2xy^2) ?

The answer to (\partial)/(\partial x)(y^2ln(y)-2xy^2) is -2y^2

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(\partial)/(\partial x)(y^2ln(y)-2xy^2)

\frac{\partial}{\partial x} (y^{2} ln (y) - 2 x y^{2})

- 2 y^{2}

Solution steps

\frac{\partial}{\partial x} (y^{2} ln (y) - 2 x y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 2 y^{2}

Simplify

= - 2 y^{2}

\frac{d}{d x} (\frac{x ^{2} - 9 x + 20}{( x + 2 ) ( x - 2 )})

\frac{d}{d x} ((lo g_{10} (x))^{s i n h (x)})

f (x) = x^{2} - 4, at a = 1

\int_{5}^{9} (9 - y + 2)^{2} d y

x \to 2 lim (\frac{1}{2} x + c)

What is (\partial)/(\partial x)(y^2ln(y)-2xy^2) ?
The answer to (\partial)/(\partial x)(y^2ln(y)-2xy^2) is -2y^2