What is (\partial)/(\partial y)(y^2ln(1+x^2+y^2)) ?

The answer to (\partial)/(\partial y)(y^2ln(1+x^2+y^2)) is 2yln(1+x^2+y^2)+(2y^3)/(1+x^2+y^2)

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(\partial)/(\partial y)(y^2ln(1+x^2+y^2))

\frac{\partial}{\partial y} (y^{2} ln (1 + x^{2} + y^{2}))

2 y ln (1 + x^{2} + y^{2}) + \frac{2 y ^{3}}{1 + x ^{2} + y ^{2}}

Solution steps

\frac{\partial}{\partial y} (y^{2} ln (1 + x^{2} + y^{2}))

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y^{2}, g = ln (1 + x^{2} + y^{2})

= \frac{\partial}{\partial y} (y^{2}) ln (1 + x^{2} + y^{2}) + \frac{\partial}{\partial y} (ln (1 + x^{2} + y^{2})) y^{2}

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

\frac{\partial}{\partial y} (ln (1 + x^{2} + y^{2})) = \frac{2 y}{1 + x ^{2} + y ^{2}}

= 2 y ln (1 + x^{2} + y^{2}) + \frac{2 y}{1 + x ^{2} + y ^{2}} y^{2}

\frac{2 y}{1 + x ^{2} + y ^{2}} y^{2} = \frac{2 y ^{3}}{1 + x ^{2} + y ^{2}}

= 2 y ln (1 + x^{2} + y^{2}) + \frac{2 y ^{3}}{1 + x ^{2} + y ^{2}}

\frac{d}{d x} (ln (\frac{x ln ( x ) - 1}{x ln ( x ) + 1}))

f (x) = \frac{( x + 1 )}{( x - 1 )}

\frac{3 t - 1}{t - 2}

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

\int \frac{2}{cos ( x )} e^{2 l n (c o s (x))} d x

What is (\partial)/(\partial y)(y^2ln(1+x^2+y^2)) ?
The answer to (\partial)/(\partial y)(y^2ln(1+x^2+y^2)) is 2yln(1+x^2+y^2)+(2y^3)/(1+x^2+y^2)