What is (\partial)/(\partial y)(x^2yz^2+sin(yz)) ?

The answer to (\partial)/(\partial y)(x^2yz^2+sin(yz)) is x^2z^2+cos(yz)z

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(\partial)/(\partial y)(x^2yz^2+sin(yz))

\frac{\partial}{\partial y} (x^{2} y z^{2} + sin (yz))

x^{2} z^{2} + cos (yz) z

Solution steps

\frac{\partial}{\partial y} (x^{2} y z^{2} + sin (yz))

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{2} y z^{2}) + \frac{\partial}{\partial y} (sin (yz))

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

\frac{\partial}{\partial y} (sin (yz)) = cos (yz) z

= x^{2} z^{2} + cos (yz) z

\frac{d x}{d t} = x - x^{2}

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

\int \frac{6 x + 4}{x ^{2} ( x ^{2} + 2 )} d x

(x - 2 y) d y + (y + 4 x) d x = 0

3 x^{4} 25 - x^{2}

What is (\partial)/(\partial y)(x^2yz^2+sin(yz)) ?
The answer to (\partial)/(\partial y)(x^2yz^2+sin(yz)) is x^2z^2+cos(yz)z