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

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

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

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

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

Solution steps

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

Treat

x, z

as constants

Apply the chain rule:

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

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

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

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

\int \frac{1}{x ^{2} x - 1} d x

\int (4^{x} - \frac{1}{3 x}) d x

f (x) = x^{7 x}

\frac{1}{1 + x + x ^{2}}

12 x^{5} - 8 x^{4} + 20 x^{3}

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