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

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

Popular >

(\partial)/(\partial x)(ln(2x^2+xy))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{1}{x 100 x ^{2} - 100} d x

f (x) = x^{9} e^{x}

y^{\frac{1}{2}} y^{^{'}} + y^{\frac{3}{2}} = 1

\frac{d}{d x} (e^{- x} sin (2 x))

x^{2} + 2 x + 1, 2 x + 5

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