What is (\partial)/(\partial x)(y*ln(x^4+y^4+1)-8) ?

The answer to (\partial)/(\partial x)(y*ln(x^4+y^4+1)-8) is (4yx^3)/(y^4+x^4+1)

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(\partial)/(\partial x)(y*ln(x^4+y^4+1)-8)

\frac{\partial}{\partial x} (y \cdot ln (x^{4} + y^{4} + 1) - 8)

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

Solution steps

\frac{\partial}{\partial x} (y ln (x^{4} + y^{4} + 1) - 8)

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (y ln (x^{4} + y^{4} + 1)) - \frac{\partial}{\partial x} (8)

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

\frac{\partial}{\partial x} (8) = 0

= \frac{4 y x ^{3}}{x ^{4} + y ^{4} + 1} - 0

Simplify

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

x \to 0 lim (\frac{( 343 + x ) ^{\frac{1}{3}} - 7}{( x )})

\int \frac{5 x + 2}{( x - 5 ) ( x + 5 ) ( x - 1 )} d x

\frac{d}{d x} (\frac{x y ^{3}}{x ^{2} + y ^{2}})

\int x^{2} + 4 x + 5 d x

\frac{\partial}{\partial x} ((9 x + 2 y)^{5})

What is (\partial)/(\partial x)(y*ln(x^4+y^4+1)-8) ?
The answer to (\partial)/(\partial x)(y*ln(x^4+y^4+1)-8) is (4yx^3)/(y^4+x^4+1)