What is the d/(dy)(ln(x^4+y^4)) ?

The d/(dy)(ln(x^4+y^4)) is (4y^3)/(x^4+y^4)

What is the first d/(dy)(ln(x^4+y^4)) ?

The first d/(dy)(ln(x^4+y^4)) is (4y^3)/(x^4+y^4)

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial w} (u (v, w) + v e^{w})

\frac{x ^{3}}{3} + \frac{1}{4 x}

f (x) = \frac{2}{5 x ^{2}}

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

y^{^{''}} - 2 y^{^{'}} - 35 y = 0

What is the d/(dy)(ln(x^4+y^4)) ?
The d/(dy)(ln(x^4+y^4)) is (4y^3)/(x^4+y^4)
What is the first d/(dy)(ln(x^4+y^4)) ?
The first d/(dy)(ln(x^4+y^4)) is (4y^3)/(x^4+y^4)