What is (\partial)/(\partial u)((x)(u+bv-x)) ?

The answer to (\partial)/(\partial u)((x)(u+bv-x)) is x

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(\partial)/(\partial u)((x)(u+bv-x))

\frac{\partial}{\partial u} ((x) (u + b v - x))

x

Solution steps

\frac{\partial}{\partial u} ((x) (u + b v - x))

Treat

b, x, v

as constants

Take the constant out:

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

= x \frac{\partial}{\partial u} (u + b v - x)

Apply the Sum/Difference Rule:

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

= x (\frac{\partial}{\partial u} (u) + \frac{\partial}{\partial u} (b v) - \frac{\partial}{\partial u} (x))

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

\frac{\partial}{\partial u} (b v) = 0

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

= x (1 + 0 - 0)

Simplify

= x

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\frac{2 y ^{3} ( y - 1 )}{( 2 y - 1 )}

\frac{e ^{- 4 x} + 3 x}{( x ) ( x ^{2} + 1 )}

π (3 r + 8 s)^{2} \cdot (12 r + 32 s)

What is (\partial)/(\partial u)((x)(u+bv-x)) ?
The answer to (\partial)/(\partial u)((x)(u+bv-x)) is x