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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x^{3} - \frac{3}{x} + e^{2})

\int (cos^{2} (x) \cdot sin (x)) d x

\int \frac{12 x ^{3}}{x ^{4} + 7} d x

y = x^{3}, at x = - 1

f (x) = \frac{2 x}{sin ( x ) + cos ( x )}

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