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}}

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

\frac{5 x + 1}{5 x - 1}

y = x, y = x^{6}

y^{^{'}} + \frac{2 y}{x} = 4 x

\int \frac{1}{4 - ( x + 1 ) ^{2}} d 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)