What is (\partial)/(\partial y)(ln(e^x+xy^3)) ?

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

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(\partial)/(\partial y)(ln(e^x+xy^3))

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

\frac{3 x y ^{2}}{e ^{x} + x y ^{3}}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{e ^{x} + x y ^{3}} \cdot 3 x y^{2} : \frac{3 x y ^{2}}{e ^{x} + x y ^{3}}

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

\frac{d}{d x} (x^{4})

f (t) = \frac{1.05}{1 + 100 e ^{- 0.285 t}}

\frac{d}{d x} (e^{\frac{1}{x} l n (x)})

(3, - 8), (12, - 15)

\frac{d y}{d x} = - \frac{x}{4 y}

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