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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Apply the chain rule:

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

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

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

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

\frac{d}{d x} (sin (x) \cdot cos (y))

n = 1 \sum \infty - \frac{2}{n}

n = 1 \sum \infty \frac{1}{n} (\frac{x}{5})^{n}

\frac{d P}{d t} = 0.1 P - 0.00013333 P^{2}

2 (ln (y)) y^{^{'}} - x^{9} y = 0

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