What is (\partial)/(\partial x)(sin((x*y)/(y-4))) ?

The answer to (\partial)/(\partial x)(sin((x*y)/(y-4))) is (ycos(\frac{xy)/(y-4))}{y-4}

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(\partial)/(\partial x)(sin((x*y)/(y-4)))

\frac{\partial}{\partial x} (sin (\frac{x \cdot y}{y - 4}))

\frac{y cos ( \frac{x y}{y - 4} )}{y - 4}

Solution steps

\frac{\partial}{\partial x} (sin (\frac{x y}{y - 4}))

Treat

y

as a constant

Apply the chain rule:

cos (\frac{x y}{y - 4}) \frac{\partial}{\partial x} (\frac{x y}{y - 4})

= cos (\frac{x y}{y - 4}) \frac{\partial}{\partial x} (\frac{x y}{y - 4})

\frac{\partial}{\partial x} (\frac{x y}{y - 4}) = \frac{y}{y - 4}

= cos (\frac{x y}{y - 4}) \frac{y}{y - 4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{y cos ( \frac{x y}{y - 4} )}{y - 4}

\int 1 + \frac{x}{x ^{2} + y ^{2}} d y

f (x) = 8 x^{2} - x^{3}, at x = 2

\frac{\partial}{\partial x} (\frac{y}{2 x})

\frac{\partial}{\partial x} (- 4 x y + x^{4} + y^{4})

y^{^{''}} + 4 y = 0

What is (\partial)/(\partial x)(sin((x*y)/(y-4))) ?
The answer to (\partial)/(\partial x)(sin((x*y)/(y-4))) is (ycos(\frac{xy)/(y-4))}{y-4}