What is (\partial)/(\partial x)(ln(7x^2+4y^2+8)) ?

The answer to (\partial)/(\partial x)(ln(7x^2+4y^2+8)) is (14x)/(7x^2+4y^2+8)

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(\partial)/(\partial x)(ln(7x^2+4y^2+8))

\frac{\partial}{\partial x} (ln (7 x^{2} + 4 y^{2} + 8))

\frac{14 x}{7 x ^{2} + 4 y ^{2} + 8}

Solution steps

\frac{\partial}{\partial x} (ln (7 x^{2} + 4 y^{2} + 8))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{7 x ^{2} + 4 y ^{2} + 8} \frac{\partial}{\partial x} (7 x^{2} + 4 y^{2} + 8)

= \frac{1}{7 x ^{2} + 4 y ^{2} + 8} \frac{\partial}{\partial x} (7 x^{2} + 4 y^{2} + 8)

\frac{\partial}{\partial x} (7 x^{2} + 4 y^{2} + 8) = 14 x

= \frac{1}{7 x ^{2} + 4 y ^{2} + 8} \cdot 14 x

Simplify

\frac{1}{7 x ^{2} + 4 y ^{2} + 8} \cdot 14 x : \frac{14 x}{7 x ^{2} + 4 y ^{2} + 8}

= \frac{14 x}{7 x ^{2} + 4 y ^{2} + 8}

y^{2} - 3 x = 4, x - y = 2

y = \frac{2}{x} - \frac{x}{2}

y = - sin (x) - \frac{1}{x}

x \to 3 + lim (5 x - [∣ x ∣])

\frac{d}{d x} ((25 - x^{2})^{\frac{1}{2}})

What is (\partial)/(\partial x)(ln(7x^2+4y^2+8)) ?
The answer to (\partial)/(\partial x)(ln(7x^2+4y^2+8)) is (14x)/(7x^2+4y^2+8)