What is (\partial)/(\partial x)(ln(x^2+y^2+1)) ?

The answer to (\partial)/(\partial x)(ln(x^2+y^2+1)) is (2x)/(x^2+y^2+1)

Popular >

(\partial)/(\partial x)(ln(x^2+y^2+1))

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

\frac{2 x}{x ^{2} + y ^{2} + 1}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} (ln (x + z))

y = 2 x - 4

\frac{d}{d x} (\frac{1}{x ^{2} + 5})

x \to 1 lim (\frac{x ^{5} - 1}{x - 1})

f (x) = x sin (2^{x})

What is (\partial)/(\partial x)(ln(x^2+y^2+1)) ?
The answer to (\partial)/(\partial x)(ln(x^2+y^2+1)) is (2x)/(x^2+y^2+1)