What is (\partial)/(\partial x)(yln(x^3+y^4+3)-9) ?

The answer to (\partial)/(\partial x)(yln(x^3+y^4+3)-9) is (3yx^2)/(y^4+x^3+3)

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(\partial)/(\partial x)(yln(x^3+y^4+3)-9)

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 3) - 9)

\frac{3 y x ^{2}}{y ^{4} + x ^{3} + 3}

Solution steps

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 3) - 9)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 3)) - \frac{\partial}{\partial x} (9)

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 3)) = \frac{3 y x ^{2}}{x ^{3} + y ^{4} + 3}

\frac{\partial}{\partial x} (9) = 0

= \frac{3 y x ^{2}}{x ^{3} + y ^{4} + 3} - 0

Simplify

= \frac{3 y x ^{2}}{y ^{4} + x ^{3} + 3}

x \to 2 lim (x^{2} - x + 2)

y^{^{'}} = t - t y

n = 0 \sum \infty \frac{10}{n}

\frac{d y}{d t} + y cos (t) = \frac{1}{100} cos (t)

\frac{\partial}{\partial x} (e^{x y z^{2}})

What is (\partial)/(\partial x)(yln(x^3+y^4+3)-9) ?
The answer to (\partial)/(\partial x)(yln(x^3+y^4+3)-9) is (3yx^2)/(y^4+x^3+3)