What is (\partial)/(\partial y)(sin(x/(1+y))) ?

The answer to (\partial)/(\partial y)(sin(x/(1+y))) is -(xcos(x/(1+y)))/((1+y)^2)

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(\partial)/(\partial y)(sin(x/(1+y)))

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 0 \sum \infty \frac{1}{n ^{6}}

\int \frac{x}{( 1 - x )} d x

\int - 4 x^{2} d x

\frac{d}{d x} (lo g_{10} (x))

f (x) = 4 e^{x} cos (x)

What is (\partial)/(\partial y)(sin(x/(1+y))) ?
The answer to (\partial)/(\partial y)(sin(x/(1+y))) is -(xcos(x/(1+y)))/((1+y)^2)