What is (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) is sec^2(6+2x^2y^3z^2)*4y^3z^2x

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(\partial)/(\partial x)(tan(6+2x^2y^3z^2))

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{3} z^{2}))

sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

Solution steps

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{3} z^{2}))

Treat

y, z

as constants

Apply the chain rule:

sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (6 + 2 x^{2} y^{3} z^{2})

= sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (6 + 2 x^{2} y^{3} z^{2})

\frac{\partial}{\partial x} (6 + 2 x^{2} y^{3} z^{2}) = 4 y^{3} z^{2} x

= sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

\frac{d}{d x} (ln (- ln (1 - x)))

2 (\frac{e ^{2 t} - e ^{2 t}}{2}) sin (2 t)

(sin (2 π x) + 2 π x cos (2 π x))^{^{'}}

\frac{d y}{d x} = \frac{x - y}{x}

\frac{\partial}{\partial x} (6 x (x^{2} + 3 y^{2} - 1)^{2})

What is (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) is sec^2(6+2x^2y^3z^2)*4y^3z^2x