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

The answer to (\partial)/(\partial x)(ln(6x^2+y^2+4)) is (12x)/(6x^2+y^2+4)

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(\partial)/(\partial x)(ln(6x^2+y^2+4))

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

\frac{12 x}{6 x ^{2} + y ^{2} + 4}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

\frac{1}{6 x ^{2} + y ^{2} + 4} \frac{\partial}{\partial x} (6 x^{2} + y^{2} + 4)

= \frac{1}{6 x ^{2} + y ^{2} + 4} \frac{\partial}{\partial x} (6 x^{2} + y^{2} + 4)

\frac{\partial}{\partial x} (6 x^{2} + y^{2} + 4) = 12 x

= \frac{1}{6 x ^{2} + y ^{2} + 4} \cdot 12 x

Simplify

\frac{1}{6 x ^{2} + y ^{2} + 4} \cdot 12 x : \frac{12 x}{6 x ^{2} + y ^{2} + 4}

= \frac{12 x}{6 x ^{2} + y ^{2} + 4}

y = x^{2} - 4 x + 4, (3, 1)

x y^{^{'}} + 2 y^{^{'}} - (x + 2) cos (x) + y = 0

n = 0 \sum \infty 2 (- \frac{1}{2})^{n}

x \to 0 lim (sin (x^{2}))

\frac{\partial}{\partial x} (2 x arctan (\frac{x}{y}))

What is (\partial)/(\partial x)(ln(6x^2+y^2+4)) ?
The answer to (\partial)/(\partial x)(ln(6x^2+y^2+4)) is (12x)/(6x^2+y^2+4)