What is (\partial)/(\partial y)(tan(7+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial y)(tan(7+2x^2y^3z^2)) is sec^2(7+2x^2y^3z^2)*6x^2z^2y^2

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(\partial)/(\partial y)(tan(7+2x^2y^3z^2))

\frac{\partial}{\partial y} (tan (7 + 2 x^{2} y^{3} z^{2}))

sec^{2} (7 + 2 x^{2} y^{3} z^{2}) \cdot 6 x^{2} z^{2} y^{2}

Solution steps

\frac{\partial}{\partial y} (tan (7 + 2 x^{2} y^{3} z^{2}))

Treat

x, z

as constants

Apply the chain rule:

sec^{2} (7 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial y} (7 + 2 x^{2} y^{3} z^{2})

= sec^{2} (7 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial y} (7 + 2 x^{2} y^{3} z^{2})

\frac{\partial}{\partial y} (7 + 2 x^{2} y^{3} z^{2}) = 6 x^{2} z^{2} y^{2}

= sec^{2} (7 + 2 x^{2} y^{3} z^{2}) \cdot 6 x^{2} z^{2} y^{2}

\frac{d}{d x} (cosh (2 x))

\int - \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

x \to 0 + lim (1 4^{x} + 2)

e^{s i n^{2} (x)}

f (x) = 7 - \frac{16}{x}, at x = 4

What is (\partial)/(\partial y)(tan(7+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial y)(tan(7+2x^2y^3z^2)) is sec^2(7+2x^2y^3z^2)*6x^2z^2y^2