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

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

Popular >

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{3}{x ^{6}})

2 cos (x) d y = (y sin (x) - y^{3}) d x

\int 3 t^{2} + 2 t d t

x \to 8 lim (\frac{3}{x - 8})

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

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