What is (\partial)/(\partial x)(ln(5x^2+8y^2+4)) ?

The answer to (\partial)/(\partial x)(ln(5x^2+8y^2+4)) is (10x)/(5x^2+8y^2+4)

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(\partial)/(\partial x)(ln(5x^2+8y^2+4))

\frac{\partial}{\partial x} (ln (5 x^{2} + 8 y^{2} + 4))

\frac{10 x}{5 x ^{2} + 8 y ^{2} + 4}

Solution steps

\frac{\partial}{\partial x} (ln (5 x^{2} + 8 y^{2} + 4))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{5 x ^{2} + 8 y ^{2} + 4} \frac{\partial}{\partial x} (5 x^{2} + 8 y^{2} + 4)

= \frac{1}{5 x ^{2} + 8 y ^{2} + 4} \frac{\partial}{\partial x} (5 x^{2} + 8 y^{2} + 4)

\frac{\partial}{\partial x} (5 x^{2} + 8 y^{2} + 4) = 10 x

= \frac{1}{5 x ^{2} + 8 y ^{2} + 4} \cdot 10 x

Simplify

\frac{1}{5 x ^{2} + 8 y ^{2} + 4} \cdot 10 x : \frac{10 x}{5 x ^{2} + 8 y ^{2} + 4}

= \frac{10 x}{5 x ^{2} + 8 y ^{2} + 4}

(\frac{x ^{2} - 1}{x ^{2} + 1})^{^{'}}

\int \frac{2}{x ^{3} - 2 x ^{2} + 9 x - 18} d x

3 e^{- x} + e^{4 x}

\frac{d}{d x} (\frac{4 x + 5}{4 x - 5})

\int (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}) d x

What is (\partial)/(\partial x)(ln(5x^2+8y^2+4)) ?
The answer to (\partial)/(\partial x)(ln(5x^2+8y^2+4)) is (10x)/(5x^2+8y^2+4)