What is the derivative of cos(2x)*ln(3x-5) ?

The derivative of cos(2x)*ln(3x-5) is -2sin(2x)ln(3x-5)+(3cos(2x))/(3x-5)

What is the first derivative of cos(2x)*ln(3x-5) ?

The first derivative of cos(2x)*ln(3x-5) is -2sin(2x)ln(3x-5)+(3cos(2x))/(3x-5)

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derivative of

cos (2 x) \cdot ln (3 x - 5)

- 2 sin (2 x) ln (3 x - 5) + \frac{3 cos ( 2 x )}{3 x - 5}

Solution steps

\frac{d}{d x} (cos (2 x) ln (3 x - 5))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = cos (2 x), g = ln (3 x - 5)

= \frac{d}{d x} (cos (2 x)) ln (3 x - 5) + \frac{d}{d x} (ln (3 x - 5)) cos (2 x)

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

\frac{d}{d x} (ln (3 x - 5)) = \frac{3}{3 x - 5}

= (- sin (2 x) \cdot 2) ln (3 x - 5) + \frac{3}{3 x - 5} cos (2 x)

Simplify

(- sin (2 x) \cdot 2) ln (3 x - 5) + \frac{3}{3 x - 5} cos (2 x) : - 2 sin (2 x) ln (3 x - 5) + \frac{3 cos ( 2 x )}{3 x - 5}

= - 2 sin (2 x) ln (3 x - 5) + \frac{3 cos ( 2 x )}{3 x - 5}

\frac{x ^{2}}{4 + x}

\frac{\partial}{\partial x} (6 e^{x^{5} y^{4}})

x \to 0 lim (\frac{x}{tan ( 7 x )})

\frac{\partial}{\partial x} (e^{- x yz})

\frac{\partial}{\partial x} (ln (x + 3))

What is the derivative of cos(2x)*ln(3x-5) ?
The derivative of cos(2x)*ln(3x-5) is -2sin(2x)ln(3x-5)+(3cos(2x))/(3x-5)
What is the first derivative of cos(2x)*ln(3x-5) ?
The first derivative of cos(2x)*ln(3x-5) is -2sin(2x)ln(3x-5)+(3cos(2x))/(3x-5)