What is (\partial)/(\partial y)((u)(Cy(2h-y))) ?

The answer to (\partial)/(\partial y)((u)(Cy(2h-y))) is uC(2h-2y)

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(\partial)/(\partial y)((u)(Cy(2h-y)))

\frac{\partial}{\partial y} ((u) (C y (2 h - y)))

u C (2 h - 2 y)

Solution steps

\frac{\partial}{\partial y} ((u) (C y (2 h - y)))

Treat

u, C, h

as constants

Take the constant out:

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

= u C \frac{\partial}{\partial y} (y (2 h - y))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = 2 h - y

= u C (\frac{\partial}{\partial y} (y) (2 h - y) + \frac{\partial}{\partial y} (2 h - y) y)

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

\frac{\partial}{\partial y} (2 h - y) = - 1

= u C (1 \cdot (2 h - y) + (- 1) y)

Simplify

u C (1 \cdot (2 h - y) + (- 1) y) : u C (2 h - 2 y)

= u C (2 h - 2 y)

\frac{( 4 - x ) ^{3}}{64}

(- x^{(2)} + 3 x + 1) d x

\frac{5 x}{2 ( x - 3 )}

(2 n - 1) (2 n +!)

\frac{( u + 3 ) ( 5 u + 1 )}{10 u ^{2} - 11 u + 1}

What is (\partial)/(\partial y)((u)(Cy(2h-y))) ?
The answer to (\partial)/(\partial y)((u)(Cy(2h-y))) is uC(2h-2y)