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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int x^{2} I n (x) d x

f (x) = 2 e^{5 x}

x \to 3 lim (x^{2} - 3 x + 5)

n = 1 \sum \infty \frac{n ^{2}}{4 ^{n}}

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

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