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

The answer to (\partial)/(\partial x)((sin(x+y))/(y^2)) is (cos(x+y))/(y^2)

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(\partial)/(\partial x)((sin(x+y))/(y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

4 \frac{d y}{d x} - 12 x^{2} = 0

y = \frac{80}{x ^{6}}

y = 3 x + 2 cos (x)

\int 6 e^{4 x} d x

\frac{\partial}{\partial x} ((x - y)^{n})

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