What is (\partial)/(\partial x)(ln(cos(x^2y^2+z))) ?

The answer to (\partial)/(\partial x)(ln(cos(x^2y^2+z))) is -2xy^2tan(x^2y^2+z)

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(\partial)/(\partial x)(ln(cos(x^2y^2+z)))

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

- 2 x y^{2} tan (x^{2} y^{2} + z)

Solution steps

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

Treat

y, z

as constants

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (cos (x^{2} y^{2} + z)) = - sin (x^{2} y^{2} + z) \cdot 2 y^{2} x

= \frac{1}{cos ( x ^{2} y ^{2} + z )} (- sin (x^{2} y^{2} + z) \cdot 2 y^{2} x)

Simplify

\frac{1}{cos ( x ^{2} y ^{2} + z )} (- sin (x^{2} y^{2} + z) \cdot 2 y^{2} x) : - 2 x y^{2} tan (x^{2} y^{2} + z)

= - 2 x y^{2} tan (x^{2} y^{2} + z)

arcsin (\frac{x}{7}), at x = 4

\frac{d}{d x} (ln (8 - 9 x))

y^{^{''}} + 4 y = t sin (2 t)

\frac{d y}{d x} + 12 y = 7

\frac{\partial}{\partial x} (\frac{x}{y} + \frac{y}{z})

What is (\partial)/(\partial x)(ln(cos(x^2y^2+z))) ?
The answer to (\partial)/(\partial x)(ln(cos(x^2y^2+z))) is -2xy^2tan(x^2y^2+z)