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

The answer to (\partial)/(\partial x)(ln(4x^2+y^2+2)) is (8x)/(4x^2+y^2+2)

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

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

\frac{8 x}{4 x ^{2} + y ^{2} + 2}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (4 x^{2} + y^{2} + 2) = 8 x

= \frac{1}{4 x ^{2} + y ^{2} + 2} \cdot 8 x

Simplify

\frac{1}{4 x ^{2} + y ^{2} + 2} \cdot 8 x : \frac{8 x}{4 x ^{2} + y ^{2} + 2}

= \frac{8 x}{4 x ^{2} + y ^{2} + 2}

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

\frac{d y}{d x} = \frac{( y - 2 )}{( y - 1 )}

\frac{d}{d x} (\frac{4}{sec ( x )})

\int_{- 1}^{5} (1 + 4 x) d x

y^{^{''}} - 4 y^{^{'}} + 9 y = x e^{x}

What is (\partial)/(\partial x)(ln(4x^2+y^2+2)) ?
The answer to (\partial)/(\partial x)(ln(4x^2+y^2+2)) is (8x)/(4x^2+y^2+2)