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

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

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

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

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

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

\frac{\partial}{\partial x} (\frac{3}{x ^{3}})

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{2} + 7})

f (x) = x (\frac{1}{x})

\int 15 d x

\frac{d}{d x} (4 x^{3} + 6 x^{2} - 5)

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