What is (\partial)/(\partial y)(ln(x^3+y^4)) ?

The answer to (\partial)/(\partial y)(ln(x^3+y^4)) is (4y^3)/(x^3+y^4)

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(\partial)/(\partial y)(ln(x^3+y^4))

\frac{\partial}{\partial y} (ln (x^{3} + y^{4}))

\frac{4 y ^{3}}{x ^{3} + y ^{4}}

Solution steps

\frac{\partial}{\partial y} (ln (x^{3} + y^{4}))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{x ^{3} + y ^{4}} \frac{\partial}{\partial y} (x^{3} + y^{4})

= \frac{1}{x ^{3} + y ^{4}} \frac{\partial}{\partial y} (x^{3} + y^{4})

\frac{\partial}{\partial y} (x^{3} + y^{4}) = 4 y^{3}

= \frac{1}{x ^{3} + y ^{4}} \cdot 4 y^{3}

Simplify

\frac{1}{x ^{3} + y ^{4}} \cdot 4 y^{3} : \frac{4 y ^{3}}{x ^{3} + y ^{4}}

= \frac{4 y ^{3}}{x ^{3} + y ^{4}}

\frac{\partial}{\partial t} (r sin (t))

f (x) = 2 x^{2} + 5, at x = 1

\frac{d}{d x} ((1 + 3 x^{4})^{5})

\int_{0}^{4} (16 - x^{2})^{2} d x

y = (2 x + 3)^{2}

What is (\partial)/(\partial y)(ln(x^3+y^4)) ?
The answer to (\partial)/(\partial y)(ln(x^3+y^4)) is (4y^3)/(x^3+y^4)