What is the d/(dz)((2cos(z))/(z+1)) ?

The d/(dz)((2cos(z))/(z+1)) is (2(-sin(z)(z+1)-cos(z)))/((z+1)^2)

What is the first d/(dz)((2cos(z))/(z+1)) ?

The first d/(dz)((2cos(z))/(z+1)) is (2(-sin(z)(z+1)-cos(z)))/((z+1)^2)

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\frac{d}{d z} (\frac{2 cos ( z )}{z + 1})

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

Solution steps

\frac{d}{d z} (\frac{2 cos ( z )}{z + 1})

Take the constant out:

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

= 2 \frac{d}{d z} (\frac{cos ( z )}{z + 1})

Apply the Quotient Rule:

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

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

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

\frac{d}{d z} (z + 1) = 1

= 2 \cdot \frac{( - sin ( z ) ) ( z + 1 ) - 1 \cdot cos ( z )}{( z + 1 ) ^{2}}

Simplify

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

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

\int_{0}^{π} 17 sin^{2} (2 x) d x

\int \frac{1}{1 + x} d x

e^{5 e^{x}}

\int (3 - 2 x)^{12} d x

\int \frac{x ^{2}}{9 - 25 x ^{2}} d x

What is the d/(dz)((2cos(z))/(z+1)) ?
The d/(dz)((2cos(z))/(z+1)) is (2(-sin(z)(z+1)-cos(z)))/((z+1)^2)
What is the first d/(dz)((2cos(z))/(z+1)) ?
The first d/(dz)((2cos(z))/(z+1)) is (2(-sin(z)(z+1)-cos(z)))/((z+1)^2)