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

The answer to (\partial)/(\partial z)(tan(1+2x^2y^3z^2)) is sec^2(1+2x^2y^3z^2)*4x^2y^3z

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

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

sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \cdot 4 x^{2} y^{3} z

Solution steps

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

Treat

x, y

as constants

Apply the chain rule:

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

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

\frac{\partial}{\partial z} (1 + 2 x^{2} y^{3} z^{2}) = 4 x^{2} y^{3} z

= sec^{2} (1 + 2 x^{2} y^{3} z^{2}) \cdot 4 x^{2} y^{3} z

\int_{- π}^{π} e^{- in x} d x

f (x) = \frac{4}{x}

\frac{d}{d x} (\frac{3 x ( 4 - x )}{5 x ^{2}})

x \to 0 lim (\frac{sin ( 3 x )}{x} + 2)

\frac{1}{x ^{4}} \frac{d y}{d x} = 19 + 15 y

What is (\partial)/(\partial z)(tan(1+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial z)(tan(1+2x^2y^3z^2)) is sec^2(1+2x^2y^3z^2)*4x^2y^3z