What is (\partial)/(\partial v)(ln(ucos(v))) ?

The answer to (\partial)/(\partial v)(ln(ucos(v))) is -(sin(v))/(cos(v))

(\partial)/(\partial v)(ln(ucos(v)))

\frac{\partial}{\partial v} (ln (u cos (v)))

- \frac{sin ( v )}{cos ( v )}

Solution steps

\frac{\partial}{\partial v} (ln (u cos (v)))

Treat

u

as a constant

Apply the chain rule:

\frac{1}{u cos ( v )} \frac{\partial}{\partial v} (u cos (v))

= \frac{1}{u cos ( v )} \frac{\partial}{\partial v} (u cos (v))

\frac{\partial}{\partial v} (u cos (v)) = - u sin (v)

= \frac{1}{u cos ( v )} (- u sin (v))

Simplify

\frac{1}{u cos ( v )} (- u sin (v)) : - \frac{sin ( v )}{cos ( v )}

= - \frac{sin ( v )}{cos ( v )}

\frac{\partial}{\partial y} (ln (3 x y) + x^{2})

\frac{\partial}{\partial y} (\frac{y}{( 2 x + 3 y )})

x \to \infty lim (- 1)

a re a 2 x^{2}, x^{4} - 2 x^{2}

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

What is (\partial)/(\partial v)(ln(ucos(v))) ?
The answer to (\partial)/(\partial v)(ln(ucos(v))) is -(sin(v))/(cos(v))