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

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

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(\partial)/(\partial y)(ln(x^3+1)+y^2)

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (ln (x^{3} + 1)) = 0

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

= 0 + 2 y

Simplify

= 2 y

\int_{- 2}^{- 1} 15 x^{- 4} d x

\frac{d}{d x} (f (x) (x^{3}))

\int \frac{2 e ^{2 x}}{2 + e ^{2 x}} d x

x \to 0 lim (e^{9 x})

x \to - 1 lim ((\frac{2}{3})^{x})

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