What is (\partial)/(\partial x)(ln(8x^2+3y^2+3)) ?

The answer to (\partial)/(\partial x)(ln(8x^2+3y^2+3)) is (16x)/(8x^2+3y^2+3)

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(\partial)/(\partial x)(ln(8x^2+3y^2+3))

\frac{\partial}{\partial x} (ln (8 x^{2} + 3 y^{2} + 3))

\frac{16 x}{8 x ^{2} + 3 y ^{2} + 3}

Solution steps

\frac{\partial}{\partial x} (ln (8 x^{2} + 3 y^{2} + 3))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{8 x ^{2} + 3 y ^{2} + 3} \frac{\partial}{\partial x} (8 x^{2} + 3 y^{2} + 3)

= \frac{1}{8 x ^{2} + 3 y ^{2} + 3} \frac{\partial}{\partial x} (8 x^{2} + 3 y^{2} + 3)

\frac{\partial}{\partial x} (8 x^{2} + 3 y^{2} + 3) = 16 x

= \frac{1}{8 x ^{2} + 3 y ^{2} + 3} \cdot 16 x

Simplify

\frac{1}{8 x ^{2} + 3 y ^{2} + 3} \cdot 16 x : \frac{16 x}{8 x ^{2} + 3 y ^{2} + 3}

= \frac{16 x}{8 x ^{2} + 3 y ^{2} + 3}

\int \frac{10 x ^{4} - 9 x ^{2}}{2 x ^{5} - 3 x ^{3} + 14} d x

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = 0

\frac{\partial}{\partial x} (\frac{5 y}{x ^{3}})

y^{^{''}} - 9 y^{^{'}} + 18 y = 5 sin (4 t), y (0) = 5, y^{^{'}} (0) = 2

x e^{- 3 x}

What is (\partial)/(\partial x)(ln(8x^2+3y^2+3)) ?
The answer to (\partial)/(\partial x)(ln(8x^2+3y^2+3)) is (16x)/(8x^2+3y^2+3)