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

The answer to (\partial)/(\partial y)(x^2sin^2(yz)) is x^2zsin(2yz)

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

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

x^{2} z sin (2 yz)

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the chain rule:

2 sin (yz) \frac{\partial}{\partial y} (sin (yz))

= 2 sin (yz) \frac{\partial}{\partial y} (sin (yz))

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

= x^{2} \cdot 2 sin (yz) cos (yz) z

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= x^{2} z sin (2 yz)

\int x^{n - 1} d x

\frac{1}{s}

\int sec^{2} (θ) + 6 d θ

\frac{x ^{2}}{1 + x}

\frac{d}{d r} (ln (r))

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