What is (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) is sec^2(6+2x^2y^3z^2)*4y^3z^2x

(\partial)/(\partial x)(tan(6+2x^2y^3z^2))

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

sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

Solution steps

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

Treat

y, z

as constants

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (6 + 2 x^{2} y^{3} z^{2}) = 4 y^{3} z^{2} x

= sec^{2} (6 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

\frac{d y}{d x} = \frac{x - y}{x}

\frac{\partial}{\partial x} (x^{2} - y^{2} + 5 x y)

m = 1 \sum \infty 3 \cdot (\frac{1}{3})

\int (x - 3)^{2} sin (x - 3) d x

\int 8 tan (8 x) d x

What is (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial x)(tan(6+2x^2y^3z^2)) is sec^2(6+2x^2y^3z^2)*4y^3z^2x