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

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

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(\partial)/(\partial x)(ln(x^3+y^2)-z)

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

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

Solution steps

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

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

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

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

Simplify

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

y = x^{8 x}

\frac{\partial}{\partial x} (e^{s i n (x + y)})

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

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