What is (\partial)/(\partial y)(tan(2+2x^2y^4z^2)) ?

The answer to (\partial)/(\partial y)(tan(2+2x^2y^4z^2)) is sec^2(2+2x^2y^4z^2)*8x^2z^2y^3

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(\partial)/(\partial y)(tan(2+2x^2y^4z^2))

\frac{\partial}{\partial y} (tan (2 + 2 x^{2} y^{4} z^{2}))

sec^{2} (2 + 2 x^{2} y^{4} z^{2}) \cdot 8 x^{2} z^{2} y^{3}

Solution steps

\frac{\partial}{\partial y} (tan (2 + 2 x^{2} y^{4} z^{2}))

Treat

x, z

as constants

Apply the chain rule:

sec^{2} (2 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial y} (2 + 2 x^{2} y^{4} z^{2})

= sec^{2} (2 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial y} (2 + 2 x^{2} y^{4} z^{2})

\frac{\partial}{\partial y} (2 + 2 x^{2} y^{4} z^{2}) = 8 x^{2} z^{2} y^{3}

= sec^{2} (2 + 2 x^{2} y^{4} z^{2}) \cdot 8 x^{2} z^{2} y^{3}

\frac{d}{d x} (\frac{x ^{2}}{10})

\int \frac{5}{x + 1} d x

\frac{d y}{d x} = 2 y^{2}

f (x) = \frac{1}{x - 1}

\int x \cdot \frac{1}{x + 1} d x

What is (\partial)/(\partial y)(tan(2+2x^2y^4z^2)) ?
The answer to (\partial)/(\partial y)(tan(2+2x^2y^4z^2)) is sec^2(2+2x^2y^4z^2)*8x^2z^2y^3