What is (\partial}{\partial x}(\frac{(x-q))/y*q) ?

The answer to (\partial}{\partial x}(\frac{(x-q))/y*q) is q/y

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(\partial}{\partial x}(\frac{(x-q))/y*q)

\frac{\partial}{\partial x} (\frac{( x - q )}{y} \cdot q)

\frac{q}{y}

Solution steps

\frac{\partial}{\partial x} (\frac{( x - q )}{y} q)

Treat

q, y

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial x} (q) = 0

= \frac{1}{y} q (1 - 0)

Simplify

\frac{1}{y} q (1 - 0) : \frac{q}{y}

= \frac{q}{y}

\int_{1}^{16} x d x

\frac{s}{s + 1.3}

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

f (z) = e^{\frac{z}{z - 1}}

\frac{d}{d x} (5 (x^{3} - 4 x)^{- 3})

What is (\partial}{\partial x}(\frac{(x-q))/y*q) ?
The answer to (\partial}{\partial x}(\frac{(x-q))/y*q) is q/y