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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int e^{t} sin (2 t) d t

y = 6 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (cos^{7} (2 x^{2} + 1))

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

x \to \infty lim (2 - x)

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