What is the derivative of (2cos(2x-1)/(2cos(x))) ?

The derivative of (2cos(2x-1)/(2cos(x))) is (-4sin(2x)cos(x)+sin(x)(2cos(2x)-1))/(2cos^2(x))

What is the first derivative of (2cos(2x-1)/(2cos(x))) ?

The first derivative of (2cos(2x-1)/(2cos(x))) is (-4sin(2x)cos(x)+sin(x)(2cos(2x)-1))/(2cos^2(x))

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\frac{d}{d x} (\frac{2 cos ( 2 x ) - 1}{2 cos ( x )})

\frac{- 4 sin ( 2 x ) cos ( x ) + sin ( x ) ( 2 cos ( 2 x ) - 1 )}{2 cos ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{2 cos ( 2 x ) - 1}{2 cos ( x )})

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 cos (2 x) - 1) = - 4 sin (2 x)

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

= \frac{1}{2} \cdot \frac{( - 4 sin ( 2 x ) ) cos ( x ) - ( - sin ( x ) ) ( 2 cos ( 2 x ) - 1 )}{( cos ( x ) ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( - 4 sin ( 2 x ) ) cos ( x ) - ( - sin ( x ) ) ( 2 cos ( 2 x ) - 1 )}{cos ^{2} ( x )} : \frac{- 4 sin ( 2 x ) cos ( x ) + sin ( x ) ( 2 cos ( 2 x ) - 1 )}{2 cos ^{2} ( x )}

= \frac{- 4 sin ( 2 x ) cos ( x ) + sin ( x ) ( 2 cos ( 2 x ) - 1 )}{2 cos ^{2} ( x )}

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 4} e^{3 t}

f (x) = (2 x - 8)^{5}

2 x^{3} + 4 x^{2} - 5 x - 3

\frac{d}{d x} (2 sin (5 x))

What is the derivative of (2cos(2x-1)/(2cos(x))) ?
The derivative of (2cos(2x-1)/(2cos(x))) is (-4sin(2x)cos(x)+sin(x)(2cos(2x)-1))/(2cos^2(x))
What is the first derivative of (2cos(2x-1)/(2cos(x))) ?
The first derivative of (2cos(2x-1)/(2cos(x))) is (-4sin(2x)cos(x)+sin(x)(2cos(2x)-1))/(2cos^2(x))