What is (\partial)/(\partial t)(xyz^2tan(yt)) ?

The answer to (\partial)/(\partial t)(xyz^2tan(yt)) is xy^2z^2sec^2(yt)

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(\partial)/(\partial t)(xyz^2tan(yt))

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

x y^{2} z^{2} sec^{2} (y t)

Solution steps

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

Treat

x, y, z

as constants

Take the constant out:

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

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

Apply the chain rule:

sec^{2} (y t) \frac{\partial}{\partial t} (y t)

= sec^{2} (y t) \frac{\partial}{\partial t} (y t)

\frac{\partial}{\partial t} (y t) = y

= x y z^{2} sec^{2} (y t) y

Simplify

x y z^{2} sec^{2} (y t) y : x y^{2} z^{2} sec^{2} (y t)

= x y^{2} z^{2} sec^{2} (y t)

x^{2} + x y - y^{2} = - 9, (3, 6)

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

F (x) = ln (2 x), y = 0, y = 2

5 sin (x), 5 cos (2 x), 0, \frac{π}{2}

\int_{- 3}^{3} \frac{1}{2} (81 - x^{4}) d x

What is (\partial)/(\partial t)(xyz^2tan(yt)) ?
The answer to (\partial)/(\partial t)(xyz^2tan(yt)) is xy^2z^2sec^2(yt)