What is the derivative of ((4z+4))/(3z^2+3z+3) ?

The derivative of ((4z+4))/(3z^2+3z+3) is -(4z(z+2))/(3(z^2+z+1)^2)

What is the first derivative of ((4z+4))/(3z^2+3z+3) ?

The first derivative of ((4z+4))/(3z^2+3z+3) is -(4z(z+2))/(3(z^2+z+1)^2)

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derivative of

\frac{( 4 z + 4 )}{3 z ^{2} + 3 z + 3}

- \frac{4 z ( z + 2 )}{3 ( z ^{2} + z + 1 ) ^{2}}

Solution steps

\frac{d}{d z} (\frac{4 z + 4}{3 z ^{2} + 3 z + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d z} ( 4 z + 4 ) ( 3 z ^{2} + 3 z + 3 ) - \frac{d}{d z} ( 3 z ^{2} + 3 z + 3 ) ( 4 z + 4 )}{( 3 z ^{2} + 3 z + 3 ) ^{2}}

\frac{d}{d z} (4 z + 4) = 4

\frac{d}{d z} (3 z^{2} + 3 z + 3) = 6 z + 3

= \frac{4 ( 3 z ^{2} + 3 z + 3 ) - ( 6 z + 3 ) ( 4 z + 4 )}{( 3 z ^{2} + 3 z + 3 ) ^{2}}

Simplify

\frac{4 ( 3 z ^{2} + 3 z + 3 ) - ( 6 z + 3 ) ( 4 z + 4 )}{( 3 z ^{2} + 3 z + 3 ) ^{2}} : - \frac{4 z ( z + 2 )}{3 ( z ^{2} + z + 1 ) ^{2}}

= - \frac{4 z ( z + 2 )}{3 ( z ^{2} + z + 1 ) ^{2}}

\frac{d y}{d x} = \frac{x y}{x ^{2} + 4}, y (0) = 6

10, x, \frac{x ^{2}}{40}

\int 2 cos (- 4 x) + 1 d x

\frac{\partial}{\partial x} (x^{2} - 4 x y + 3 y^{2})

\frac{d}{d x} ((x^{3} + 4) e^{x})

What is the derivative of ((4z+4))/(3z^2+3z+3) ?
The derivative of ((4z+4))/(3z^2+3z+3) is -(4z(z+2))/(3(z^2+z+1)^2)
What is the first derivative of ((4z+4))/(3z^2+3z+3) ?
The first derivative of ((4z+4))/(3z^2+3z+3) is -(4z(z+2))/(3(z^2+z+1)^2)