What is (\partial)/(\partial z)(tan(5+2x^2y^4z^2)) ?

The answer to (\partial)/(\partial z)(tan(5+2x^2y^4z^2)) is sec^2(5+2x^2y^4z^2)*4x^2y^4z

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(\partial)/(\partial z)(tan(5+2x^2y^4z^2))

\frac{\partial}{\partial z} (tan (5 + 2 x^{2} y^{4} z^{2}))

sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \cdot 4 x^{2} y^{4} z

Solution steps

\frac{\partial}{\partial z} (tan (5 + 2 x^{2} y^{4} z^{2}))

Treat

x, y

as constants

Apply the chain rule:

sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial z} (5 + 2 x^{2} y^{4} z^{2})

= sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial z} (5 + 2 x^{2} y^{4} z^{2})

\frac{\partial}{\partial z} (5 + 2 x^{2} y^{4} z^{2}) = 4 x^{2} y^{4} z

= sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \cdot 4 x^{2} y^{4} z

\frac{d y}{d x} = 5^{t a n (5 x)}

y^{^{''}} - \frac{2}{t ^{2}} y = 0

\frac{\partial}{\partial x} (x^{3} sin (x^{2}))

\int_{1}^{2} \frac{9}{x ^{3} + 2 x} d x

y^{^{''}} + 5 y^{^{'}} + 6 y = 0

What is (\partial)/(\partial z)(tan(5+2x^2y^4z^2)) ?
The answer to (\partial)/(\partial z)(tan(5+2x^2y^4z^2)) is sec^2(5+2x^2y^4z^2)*4x^2y^4z