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

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

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

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

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

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

\frac{16}{( x ^{2} y ^{2} )}

\frac{d}{d y} (ln (x - 2 y))

\frac{d}{d r} (π r^{2})

x sin^{2} (2 x)

\frac{d}{d x} (7^{x^{4} + 2})

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