What is (\partial)/(\partial y)(-ln(cos(y))) ?

The answer to (\partial)/(\partial y)(-ln(cos(y))) is tan(y)

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(\partial)/(\partial y)(-ln(cos(y)))

\frac{\partial}{\partial y} (- ln (cos (y)))

tan (y)

Solution steps

\frac{\partial}{\partial y} (- ln (cos (y)))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{\partial}{\partial y} (ln (cos (y)))

Apply the chain rule:

\frac{1}{cos ( y )} \frac{\partial}{\partial y} (cos (y))

= \frac{1}{cos ( y )} \frac{\partial}{\partial y} (cos (y))

\frac{\partial}{\partial y} (cos (y)) = - sin (y)

= - \frac{1}{cos ( y )} (- sin (y))

Simplify

- \frac{1}{cos ( y )} (- sin (y)) : tan (y)

= tan (y)

\int e^{6 x} sin (5 x) d x

x \to 0 lim (- 1 + 2)

\int 6 x^{2} 2 x^{3} + 1 d x

\frac{d y}{d t} = 4 y, y (5) = 9

What is (\partial)/(\partial y)(-ln(cos(y))) ?
The answer to (\partial)/(\partial y)(-ln(cos(y))) is tan(y)