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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (13.12 + 0.6215 x - 11.37 y^{0.16} + 0.3965 x y^{0.16})

f (x) = sin (x) cos (x)

\int \frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} d x

x \to 0 lim (8 x csc (5 x))

11 (t + 1) \frac{d y}{d t} - 8 y = 24 t

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