What is the (cos(ln(x)))^' ?

The (cos(ln(x)))^' is -(sin(ln(x)))/x

What is the first (cos(ln(x)))^' ?

The first (cos(ln(x)))^' is -(sin(ln(x)))/x

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(cos (ln (x)))^{^{'}}

- \frac{sin ( ln ( x ) )}{x}

Solution steps

(cos (ln (x)))^{'}

Apply the chain rule:

- sin (ln (x)) (ln (x))^{'}

= - sin (ln (x)) (ln (x))^{'}

(ln (x))^{'} = \frac{1}{x}

= - sin (ln (x)) \frac{1}{x}

Simplify

- sin (ln (x)) \frac{1}{x} : - \frac{sin ( ln ( x ) )}{x}

= - \frac{sin ( ln ( x ) )}{x}

\frac{d}{d x} (a r (x) c)

2 sin (x) + sin^{2} (x)

\frac{\partial}{\partial t} (y + x cos (t))

\int 9 x^{2} d x

y^{^{''}} + 14 y^{^{'}} + 50 y = 0, y (0) = 2, y^{^{'}} (0) = - 17