What is (\partial)/(\partial y)((z)(y^2z+yln(z)-z^2e^{xz}-3)) ?

The answer to (\partial)/(\partial y)((z)(y^2z+yln(z)-z^2e^{xz}-3)) is z(2zy+ln(z))

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(\partial)/(\partial y)((z)(y^2z+yln(z)-z^2e^{xz}-3))

\frac{\partial}{\partial y} ((z) (y^{2} z + y ln (z) - z^{2} e^{x z} - 3))

z (2 zy + ln (z))

Solution steps

\frac{\partial}{\partial y} ((z) (y^{2} z + y ln (z) - z^{2} e^{x z} - 3))

Treat

z, x

as constants

Take the constant out:

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

= z \frac{\partial}{\partial y} (y^{2} z + y ln (z) - z^{2} e^{x z} - 3)

Apply the Sum/Difference Rule:

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

= z (\frac{\partial}{\partial y} (y^{2} z) + \frac{\partial}{\partial y} (y ln (z)) - \frac{\partial}{\partial y} (z^{2} e^{x z}) - \frac{\partial}{\partial y} (3))

\frac{\partial}{\partial y} (y^{2} z) = 2 zy

\frac{\partial}{\partial y} (y ln (z)) = ln (z)

\frac{\partial}{\partial y} (z^{2} e^{x z}) = 0

\frac{\partial}{\partial y} (3) = 0

= z (2 zy + ln (z) - 0 - 0)

Simplify

= z (2 zy + ln (z))

\frac{( 1 + i ) ^{6}}{( 1 - i ) ^{2} ( 2 + i ) ^{4}}

2 (n (m + 1))

\frac{1}{( s + 1 ) \cdot ( s + 2 )}

(x^{\frac{1}{2}} + x^{\frac{2}{3}})^{9}

a (4, - 5) b (10, 4)

What is (\partial)/(\partial y)((z)(y^2z+yln(z)-z^2e^{xz}-3)) ?
The answer to (\partial)/(\partial y)((z)(y^2z+yln(z)-z^2e^{xz}-3)) is z(2zy+ln(z))