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

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

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

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

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

Solution steps

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

Treat

y, z

as constants

Apply the chain rule:

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

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

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

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

\int (- 3 cos (x) + 4 sec^{2} (x)) d x

\frac{d}{d x} (y e^{2 x})

f (x) = \frac{x}{7}

x \to 0 lim (\frac{sin ^{4} ( 2 x )}{4 x ^{4}})

\frac{d v}{d t} = 0.64 - \frac{v}{2940} \cdot 8, v (0) = 0

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