What is (\partial)/(\partial x)(ln(7x^2+xy)) ?

The answer to (\partial)/(\partial x)(ln(7x^2+xy)) is (14x+y)/(7x^2+xy)

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(\partial)/(\partial x)(ln(7x^2+xy))

\frac{\partial}{\partial x} (ln (7 x^{2} + x y))

\frac{14 x + y}{7 x ^{2} + x y}

Solution steps

\frac{\partial}{\partial x} (ln (7 x^{2} + x y))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{7 x ^{2} + x y} \frac{\partial}{\partial x} (7 x^{2} + x y)

= \frac{1}{7 x ^{2} + x y} \frac{\partial}{\partial x} (7 x^{2} + x y)

\frac{\partial}{\partial x} (7 x^{2} + x y) = 14 x + y

= \frac{1}{7 x ^{2} + x y} (14 x + y)

Simplify

\frac{1}{7 x ^{2} + x y} (14 x + y) : \frac{14 x + y}{7 x ^{2} + x y}

= \frac{14 x + y}{7 x ^{2} + x y}

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (e x)

\int \frac{9 x ^{2}}{2} - 5 cos (x) + 7 d x

- \frac{1}{s ^{2}}

\int e^{- c x} d x

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

What is (\partial)/(\partial x)(ln(7x^2+xy)) ?
The answer to (\partial)/(\partial x)(ln(7x^2+xy)) is (14x+y)/(7x^2+xy)