What is (\partial)/(\partial x)(tan(9+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial x)(tan(9+2x^2y^3z^2)) is sec^2(9+2x^2y^3z^2)*4y^3z^2x

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(\partial)/(\partial x)(tan(9+2x^2y^3z^2))

\frac{\partial}{\partial x} (tan (9 + 2 x^{2} y^{3} z^{2}))

sec^{2} (9 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

Solution steps

\frac{\partial}{\partial x} (tan (9 + 2 x^{2} y^{3} z^{2}))

Treat

y, z

as constants

Apply the chain rule:

sec^{2} (9 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (9 + 2 x^{2} y^{3} z^{2})

= sec^{2} (9 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (9 + 2 x^{2} y^{3} z^{2})

\frac{\partial}{\partial x} (9 + 2 x^{2} y^{3} z^{2}) = 4 y^{3} z^{2} x

= sec^{2} (9 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

\frac{\partial}{\partial x} (ln (z))

x \to 0 lim (1 - \frac{tan ( k x )}{x})

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

\frac{d}{d x} (\frac{( ln ( x ) ) ^{2}}{x})

x y^{^{'}} + 2 y = 6 x^{2} y

What is (\partial)/(\partial x)(tan(9+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial x)(tan(9+2x^2y^3z^2)) is sec^2(9+2x^2y^3z^2)*4y^3z^2x