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}

\frac{d}{d x} (sec (x) (x - tan (x)))

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x}

x = 6 y^{2}, x = 4 + 2 y^{2}

f (x) = x (x^{6} - 1), at x = 1

\int e^{x} \cdot cos (3 x) d x

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)