What is (\partial)/(\partial y)(tan(5+2x^2y^4z^2)) ?

The answer to (\partial)/(\partial y)(tan(5+2x^2y^4z^2)) is sec^2(5+2x^2y^4z^2)*8x^2z^2y^3

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(\partial)/(\partial y)(tan(5+2x^2y^4z^2))

\frac{\partial}{\partial y} (tan (5 + 2 x^{2} y^{4} z^{2}))

sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \cdot 8 x^{2} z^{2} y^{3}

Solution steps

\frac{\partial}{\partial y} (tan (5 + 2 x^{2} y^{4} z^{2}))

Treat

x, z

as constants

Apply the chain rule:

sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial y} (5 + 2 x^{2} y^{4} z^{2})

= sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \frac{\partial}{\partial y} (5 + 2 x^{2} y^{4} z^{2})

\frac{\partial}{\partial y} (5 + 2 x^{2} y^{4} z^{2}) = 8 x^{2} z^{2} y^{3}

= sec^{2} (5 + 2 x^{2} y^{4} z^{2}) \cdot 8 x^{2} z^{2} y^{3}

\frac{\partial}{\partial y} (e^{y} x)

\int cos (θ) sin^{3} (θ) d θ

9 y^{^{''}} + 32 y = 0, y^{^{'}} (0) = 9, y (0) = 5

(x - 8 y) d x = - 4 x d y

\int 1 + 3 cos^{2} (x) sin (2 x) d x

What is (\partial)/(\partial y)(tan(5+2x^2y^4z^2)) ?
The answer to (\partial)/(\partial y)(tan(5+2x^2y^4z^2)) is sec^2(5+2x^2y^4z^2)*8x^2z^2y^3