What is (\partial)/(\partial x)(tan(3+2x^2y^4z^2)) ?

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

Popular >

(\partial)/(\partial x)(tan(3+2x^2y^4z^2))

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

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

Solution steps

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

Treat

y, z

as constants

Apply the chain rule:

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

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

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

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

\int - x e^{2 x} d x

\int 9 x arcsin (x) d x

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

x \to 0 lim (\frac{∣ x - 3 ∣}{x - 3})

\frac{\partial}{\partial x} (ln (x^{6} + y^{6}))

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