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

The answer to (\partial)/(\partial x)(ln(3x^2+y^2+4)) is (6x)/(3x^2+y^2+4)

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

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

\frac{6 x}{3 x ^{2} + y ^{2} + 4}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (3 x^{2} + y^{2} + 4) = 6 x

= \frac{1}{3 x ^{2} + y ^{2} + 4} \cdot 6 x

Simplify

\frac{1}{3 x ^{2} + y ^{2} + 4} \cdot 6 x : \frac{6 x}{3 x ^{2} + y ^{2} + 4}

= \frac{6 x}{3 x ^{2} + y ^{2} + 4}

What is (\partial)/(\partial x)(ln(3x^2+y^2+4)) ?
The answer to (\partial)/(\partial x)(ln(3x^2+y^2+4)) is (6x)/(3x^2+y^2+4)