What is the derivative of (ln(cos(2x)))/(5x) ?

The derivative of (ln(cos(2x)))/(5x) is (-2xtan(2x)-ln(cos(2x)))/(5x^2)

What is the first derivative of (ln(cos(2x)))/(5x) ?

The first derivative of (ln(cos(2x)))/(5x) is (-2xtan(2x)-ln(cos(2x)))/(5x^2)

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derivative of

\frac{ln ( cos ( 2 x ) )}{5 x}

\frac{- 2 x tan ( 2 x ) - ln ( cos ( 2 x ) )}{5 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{ln ( cos ( 2 x ) )}{5 x})

Take the constant out:

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

= \frac{1}{5} \frac{d}{d x} (\frac{ln ( cos ( 2 x ) )}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{5} \cdot \frac{\frac{d}{d x} ( ln ( cos ( 2 x ) ) ) x - \frac{d x}{d x} ln ( cos ( 2 x ) )}{x ^{2}}

\frac{d}{d x} (ln (cos (2 x))) = - 2 tan (2 x)

\frac{d x}{d x} = 1

= \frac{1}{5} \cdot \frac{( - 2 tan ( 2 x ) ) x - 1 \cdot ln ( cos ( 2 x ) )}{x ^{2}}

Simplify

\frac{1}{5} \cdot \frac{( - 2 tan ( 2 x ) ) x - 1 \cdot ln ( cos ( 2 x ) )}{x ^{2}} : \frac{- 2 x tan ( 2 x ) - ln ( cos ( 2 x ) )}{5 x ^{2}}

= \frac{- 2 x tan ( 2 x ) - ln ( cos ( 2 x ) )}{5 x ^{2}}

\frac{( x - 9 )}{( 5 x - 1 )}, at x = 1

(x - 6)^{2}

6 y^{^{''}} - 54 y = 0, y (- 2) = - 1, y^{^{'}} (2) = 1

x \to 0 lim (\frac{ln ( x )}{x ^{- 2}})

f (t) = t e^{- 1} sin (t)

What is the derivative of (ln(cos(2x)))/(5x) ?
The derivative of (ln(cos(2x)))/(5x) is (-2xtan(2x)-ln(cos(2x)))/(5x^2)
What is the first derivative of (ln(cos(2x)))/(5x) ?
The first derivative of (ln(cos(2x)))/(5x) is (-2xtan(2x)-ln(cos(2x)))/(5x^2)