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

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

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

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

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

Solution steps

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

Treat

y

as a constant

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

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

Simplify

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

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

\int \frac{x ^{2}}{1 - x ^{6}} d x

\int \frac{2 x - 5}{( x + 2 ) ( x + 3 )} d x

\int x cos (x) d x

\frac{3 x ^{\frac{1}{2}}}{2}

x \to - 7 lim (\frac{6 x + 42}{x + 7})

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