What is (\partial)/(\partial x)(kln(sqrt(x^2+y^2))) ?

The answer to (\partial)/(\partial x)(kln(sqrt(x^2+y^2))) is (kx)/(x^2+y^2)

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(\partial)/(\partial x)(kln(sqrt(x^2+y^2)))

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

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

Solution steps

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

Treat

k, y

as a constant

Take the constant out:

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

= k \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}) = \frac{x}{x ^{2} + y ^{2}}

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

Simplify

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

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

n = 1 \sum \infty (- \frac{1}{2})^{n + 1}

\int cos^{5} (x) tan^{4} (x) d x

\frac{d y}{d x} (1 + e^{x}) = e^{x - y}

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\frac{\partial}{\partial x} (2 x - y e^{y})

What is (\partial)/(\partial x)(kln(sqrt(x^2+y^2))) ?
The answer to (\partial)/(\partial x)(kln(sqrt(x^2+y^2))) is (kx)/(x^2+y^2)