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

The answer to (\partial)/(\partial x)(ln(x^3+y)-z) is (3x^2)/(x^3+y)

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(\partial)/(\partial x)(ln(x^3+y)-z)

\frac{\partial}{\partial x} (ln (x^{3} + y) - z)

\frac{3 x ^{2}}{x ^{3} + y}

Solution steps

\frac{\partial}{\partial x} (ln (x^{3} + y) - z)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (ln (x^{3} + y)) = \frac{3 x ^{2}}{x ^{3} + y}

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

= \frac{3 x ^{2}}{x ^{3} + y} - 0

Simplify

= \frac{3 x ^{2}}{x ^{3} + y}

\int \frac{x ^{10}}{( x ^{2} + x - 2 )} d x

\frac{d}{d x} (e^{- \frac{1}{2}})

f (x) = \frac{1}{2} x^{2} - 2 x + 6

x \cdot \frac{d y}{d x} = 2 y

\int (csc (x) - tan (x))^{2} d x

What is (\partial)/(\partial x)(ln(x^3+y)-z) ?
The answer to (\partial)/(\partial x)(ln(x^3+y)-z) is (3x^2)/(x^3+y)