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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

= 7 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x

Simplify

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

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

n = 5 \sum \infty \frac{1}{n}

x^{^{''}} - 5 x^{^{'}} + 6 x = e^{t}

\frac{d}{d x} (- \frac{7}{x ^{8}})

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

\frac{d}{d x} (x^{\frac{1}{3}} (x + 3)^{\frac{2}{3}})

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