What is (\partial)/(\partial x)(-(xcos(y))/(x-1)) ?

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

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(\partial)/(\partial x)(-(xcos(y))/(x-1))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{\partial}{\partial x} (x - 1) = 1

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

Simplify

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

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

sin (tan (6 x))

n = 1 \sum \infty n^{- 8}

ln (\frac{q}{7})

\frac{\partial}{\partial x} (4 x e^{- x^{2} - y^{2} - z^{2}})

\int_{6}^{8} \frac{36}{( x - 6 ) ^{3}} d x

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