What is (\partial)/(\partial y)(tan(1+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial y)(tan(1+2x^2y^3z^2)) is sec^2(1+2x^2y^3z^2)*6x^2z^2y^2

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(\partial)/(\partial y)(tan(1+2x^2y^3z^2))

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{3} z^{2}))

sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \cdot 6 x^{2} z^{2} y^{2}

Solution steps

\frac{\partial}{\partial y} (tan (1 + 2 x^{2} y^{3} z^{2}))

Treat

x, z

as constants

Apply the chain rule:

sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial y} (1 + 2 x^{2} y^{3} z^{2})

= sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial y} (1 + 2 x^{2} y^{3} z^{2})

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

= sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \cdot 6 x^{2} z^{2} y^{2}

(x + \frac{1}{x})^{^{'}}

[x + (x + sin^{2} (x))^{3}]^{5}

\frac{d}{d x} (\frac{ln ( x + 1 )}{4 x + 2})

\int sec^{3} (4 x) d x

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

What is (\partial)/(\partial y)(tan(1+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial y)(tan(1+2x^2y^3z^2)) is sec^2(1+2x^2y^3z^2)*6x^2z^2y^2