What is (\partial)/(\partial u)((y)(uv-z(y,u,v))) ?

The answer to (\partial)/(\partial u)((y)(uv-z(y,u,v))) is y(v-z(y,,v*1))

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(\partial)/(\partial u)((y)(uv-z(y,u,v)))

\frac{\partial}{\partial u} ((y) (uv - z (y, u, v)))

y (v - z (y,, v \cdot 1))

Solution steps

\frac{\partial}{\partial u} ((y) (uv - z (y, u, v)))

Treat

y, v, z

as constants

Take the constant out:

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

= y \frac{\partial}{\partial u} (uv - z (y, u, v))

Apply the Sum/Difference Rule:

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

= y (\frac{\partial}{\partial u} (uv) - \frac{\partial}{\partial u} (z (y, u, v)))

\frac{\partial}{\partial u} (uv) = v

\frac{\partial}{\partial u} (z (y, u, v)) = z (y,, v \cdot 1)

= y (v - z (y,, v \cdot 1))

\frac{( - 2 x - 3 x ^{2} ) ( 4 - x ^{2} - x ^{3} )}{h}

3 sin (\frac{( π ( x - 5 ) )}{( 6 )})

\frac{1}{( j x + 4 ) \cdot ( j x + 1 )}

\frac{\partial ^{2}}{\partial u \partial x} ((x) (- 2 x + u (x)))

\frac{( A s ) + B}{( s ^{2} + 1 ) ^{2}} + \frac{( C s ) + D}{s ^{2} + 1}

What is (\partial)/(\partial u)((y)(uv-z(y,u,v))) ?
The answer to (\partial)/(\partial u)((y)(uv-z(y,u,v))) is y(v-z(y,,v*1))