What is (\partial)/(\partial k)(ln(sin(x/k))) ?

The answer to (\partial)/(\partial k)(ln(sin(x/k))) is -(xcot(x/k))/(k^2)

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(\partial)/(\partial k)(ln(sin(x/k)))

\frac{\partial}{\partial k} (ln (sin (\frac{x}{k})))

- \frac{x cot ( \frac{x}{k} )}{k ^{2}}

Solution steps

\frac{\partial}{\partial k} (ln (sin (\frac{x}{k})))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{sin ( \frac{x}{k} )} \frac{\partial}{\partial k} (sin (\frac{x}{k}))

= \frac{1}{sin ( \frac{x}{k} )} \frac{\partial}{\partial k} (sin (\frac{x}{k}))

\frac{\partial}{\partial k} (sin (\frac{x}{k})) = - \frac{x cos ( \frac{x}{k} )}{k ^{2}}

= \frac{1}{sin ( \frac{x}{k} )} (- \frac{x cos ( \frac{x}{k} )}{k ^{2}})

Simplify

\frac{1}{sin ( \frac{x}{k} )} (- \frac{x cos ( \frac{x}{k} )}{k ^{2}}) : - \frac{x cot ( \frac{x}{k} )}{k ^{2}}

= - \frac{x cot ( \frac{x}{k} )}{k ^{2}}

\int \frac{x ^{2}}{4 x ^{3} + 7} d x

\frac{d}{d x} (\frac{1}{3} tan^{3} (\frac{1}{x}))

(x^{2} - x) y^{^{'}} = y - 2 x y

cos^{4} (θ + 5)

n = 1 \sum \infty \frac{n}{n ^{3} + 1}

What is (\partial)/(\partial k)(ln(sin(x/k))) ?
The answer to (\partial)/(\partial k)(ln(sin(x/k))) is -(xcot(x/k))/(k^2)