What is (\partial)/(\partial x)(yln(x^3+y^4+2)) ?

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

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(\partial)/(\partial x)(yln(x^3+y^4+2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

\frac{1}{x ^{3} + y ^{4} + 2} \frac{\partial}{\partial x} (x^{3} + y^{4} + 2)

= \frac{1}{x ^{3} + y ^{4} + 2} \frac{\partial}{\partial x} (x^{3} + y^{4} + 2)

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

= y \frac{1}{x ^{3} + y ^{4} + 2} \cdot 3 x^{2}

Simplify

y \frac{1}{x ^{3} + y ^{4} + 2} \cdot 3 x^{2} : \frac{3 y x ^{2}}{x ^{3} + y ^{4} + 2}

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

\frac{d}{d x} (e^{x z (x)})

f (x) = x^{2} + 169 x + 820, at x = 2

y = \frac{2 x}{7 - cot ( x )}

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

\int_{1}^{9} (x) d x

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