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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} = y (2 - t y)

\frac{\partial}{\partial x} (2 x y^{2} z - 4 x z^{2})

y = \frac{3 ( x - 1 )}{4 x x}

\int x sin (\frac{nπ x}{10}) d x

f (t) = t^{9} - 1

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