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

The answer to (\partial)/(\partial x)(ln(x+y^4)-z) is 1/(x+y^4)

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

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

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

Solution steps

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

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

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

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

Simplify

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

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What is (\partial)/(\partial x)(ln(x+y^4)-z) ?
The answer to (\partial)/(\partial x)(ln(x+y^4)-z) is 1/(x+y^4)