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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (- 8 sin (\frac{3 x}{2} + 1))

x \to 4 - lim (2 x - 1)

f (x) = \frac{( x - 4 )}{x - 2}

\int_{3}^{5} 1

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