What is (\partial)/(\partial x)(ln(x^2+y^4)-z) ?

The answer to (\partial)/(\partial x)(ln(x^2+y^4)-z) is (2x)/(x^2+y^4)

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(\partial)/(\partial x)(ln(x^2+y^4)-z)

\frac{\partial}{\partial x} (ln (x^{2} + y^{4}) - z)

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

Solution steps

\frac{\partial}{\partial x} (ln (x^{2} + 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^{2} + y^{4})) - \frac{\partial}{\partial x} (z)

\frac{\partial}{\partial x} (ln (x^{2} + y^{4})) = \frac{2 x}{x ^{2} + y ^{4}}

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

= \frac{2 x}{x ^{2} + y ^{4}} - 0

Simplify

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

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

\frac{d}{d x} (2 e^{- x^{2}})

\int 6 cos (3 x) d x

\frac{d}{d x} (a cos (2 x) + b sin (2 x))

\frac{d}{d x} (a sin (x))

What is (\partial)/(\partial x)(ln(x^2+y^4)-z) ?
The answer to (\partial)/(\partial x)(ln(x^2+y^4)-z) is (2x)/(x^2+y^4)